{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0001","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"681649bbdbb26bf0376cf0bc0eb24e1d9c38c3d8f6fb9dc19e6327febf79bb5f","family_id":"qf-str22-oscillator-model-qualification","seed":202610221001,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[-1,1],\"spring\":[-8,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=-1/1 and spring coefficient k=-8/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(-1, -8, No, No, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (-1, -8, No, No, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0002","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d0f30f4fbec6ee81619ef36c8aacc7b39f0da209e50025c372fad1575f26da9a","family_id":"qf-str22-oscillator-model-qualification","seed":202610221002,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[0,1],\"spring\":[-7,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=0/1 and spring coefficient k=-7/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(0, -7, No, No, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (0, -7, No, No, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0003","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19d2bc7640e627981e4d8cb8a9f2fd37172af4862e5f5f5403007ba31da189a5","family_id":"qf-str22-oscillator-model-qualification","seed":202610221003,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[1,1],\"spring\":[-6,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=-6/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(1, -6, Yes, No, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, -6, Yes, No, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0004","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49cb228e326870df14147280e5e1697652ee1872887739eac730b75f3bea5c62","family_id":"qf-str22-oscillator-model-qualification","seed":202610221004,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[2,1],\"spring\":[-5,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=-5/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(2, -5, Yes, No, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, -5, Yes, No, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0005","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a950b46c5101312ea6b97adaf5fc98ce9a8d032035ca74e21994ceb6abf94237","family_id":"qf-str22-oscillator-model-qualification","seed":202610221005,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[3,1],\"spring\":[-4,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=-4/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(3, -4, Yes, No, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, -4, Yes, No, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0006","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"20655b12654823667ae5c01467463925266c409187084b693b7b030935f7fab2","family_id":"qf-str22-oscillator-model-qualification","seed":202610221006,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[-1,1],\"spring\":[-3,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=-1/1 and spring coefficient k=-3/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(-1, -3, No, No, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (-1, -3, No, No, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0007","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f56cbf7837d031da166472805835ef128df9aaef121ec32b8860a9592efc1ff","family_id":"qf-str22-oscillator-model-qualification","seed":202610221007,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[0,1],\"spring\":[-2,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=0/1 and spring coefficient k=-2/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(0, -2, No, No, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (0, -2, No, No, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0008","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"252f3293fd4c2340a0e43f61fe6940ee5a0cb7843155b456ba0e045212fa07c9","family_id":"qf-str22-oscillator-model-qualification","seed":202610221008,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[1,1],\"spring\":[-1,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=-1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(1, -1, Yes, No, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, -1, Yes, No, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0009","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f1f503741f63e8275a3507d4141d3d5d1ac7b601d1df803117ff369eba276923","family_id":"qf-str22-oscillator-model-qualification","seed":202610221009,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[2,1],\"spring\":[0,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=0/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(2, 0, Yes, No, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 0, Yes, No, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0010","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"824986da226bc562591f075d1cff41a7dde2edd4b8d60712b40c3760a1e279c4","family_id":"qf-str22-oscillator-model-qualification","seed":202610221010,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[3,1],\"spring\":[1,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(3, 1, Yes, Yes, Yes)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 1, Yes, Yes, Yes)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0011","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a0dc31081519fdb20a65c4b1d5396b11aabaf0e7cbc113fb1d1d4cecaecc7d6","family_id":"qf-str22-oscillator-model-qualification","seed":202610221011,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[-1,1],\"spring\":[2,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=-1/1 and spring coefficient k=2/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(-1, 2, No, Yes, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (-1, 2, No, Yes, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0012","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"62effb21edd2b49e3f5018a1c5aa5b77593985a89fe9fc6c155d99bd3c1eb9c6","family_id":"qf-str22-oscillator-model-qualification","seed":202610221012,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[0,1],\"spring\":[3,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=0/1 and spring coefficient k=3/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(0, 3, No, Yes, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (0, 3, No, Yes, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0013","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3c5aaddab8393ba53df7fc10acd60183efe99d1bb9f4891107c96579499a467","family_id":"qf-str22-oscillator-model-qualification","seed":202610221013,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[1,1],\"spring\":[4,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=4/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(1, 4, Yes, Yes, Yes)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 4, Yes, Yes, Yes)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0014","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"85a47674e8971f4a1d130c034b418f287fc236d608b64b869854c2cfb7c898b3","family_id":"qf-str22-oscillator-model-qualification","seed":202610221014,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[2,1],\"spring\":[5,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=5/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(2, 5, Yes, Yes, Yes)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 5, Yes, Yes, Yes)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0015","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2deb3e1ffe96fc12d626349594964860afaac0217a53485f04ebae174a36476b","family_id":"qf-str22-oscillator-model-qualification","seed":202610221015,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[3,1],\"spring\":[6,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=6/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(3, 6, Yes, Yes, Yes)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 6, Yes, Yes, Yes)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0016","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e73a954d9470c3fde8b93c3da377cd9add7fc97f71f51b122d73df0dd3cdc4f0","family_id":"qf-str22-oscillator-model-qualification","seed":202610221016,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[-1,1],\"spring\":[7,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=-1/1 and spring coefficient k=7/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(-1, 7, No, Yes, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (-1, 7, No, Yes, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0017","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a835e46d14c6df2a3f3e102a1ce68378b104a88d0b6795c54027011869e94fb3","family_id":"qf-str22-oscillator-model-qualification","seed":202610221017,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[0,1],\"spring\":[8,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=0/1 and spring coefficient k=8/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(0, 8, No, Yes, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (0, 8, No, Yes, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0018","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09e03bad6196b50dd840e62c193c65c0ab0765e4b7fd0813837a75da1810dd61","family_id":"qf-str22-oscillator-model-qualification","seed":202610221018,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[1,1],\"spring\":[9,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=9/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(1, 9, Yes, Yes, Yes)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 9, Yes, Yes, Yes)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0019","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5efdf0a926ccd2d67ad1c8a66bd07a576b0cf0e9fdb6d691ec63ade40434cc93","family_id":"qf-str22-oscillator-model-qualification","seed":202610221019,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[2,1],\"spring\":[10,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=10/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(2, 10, Yes, Yes, Yes)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 10, Yes, Yes, Yes)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0020","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e24832864fa065436773384d4b8e0ea12e088782a8f03bf321081a64bbe5805","family_id":"qf-str22-oscillator-model-qualification","seed":202610221020,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[3,1],\"spring\":[11,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=11/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(3, 11, Yes, Yes, Yes)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 11, Yes, Yes, Yes)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0021","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be04a28711450d9f3c897b781a4dee58f170481110e7d9d017171e62ea6f0075","family_id":"qf-str22-oscillator-model-qualification","seed":202610221021,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[-1,1],\"spring\":[12,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=-1/1 and spring coefficient k=12/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(-1, 12, No, Yes, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (-1, 12, No, Yes, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0022","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9dd2065a52a56f4fc630114ff460781f0f202af4db3fd01294702892496093a7","family_id":"qf-str22-oscillator-model-qualification","seed":202610221022,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[0,1],\"spring\":[13,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=0/1 and spring coefficient k=13/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(0, 13, No, Yes, No)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (0, 13, No, Yes, No)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0023","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"76d72dc53b347c61bc1e84f83c14ca4de06d84d7a5e19b043fe49668b8c40639","family_id":"qf-str22-oscillator-model-qualification","seed":202610221023,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[1,1],\"spring\":[14,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=14/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(1, 14, Yes, Yes, Yes)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 14, Yes, Yes, Yes)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0024","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fdcbb87bd65f6e5d95e1769f747c12c8d85a37f58cbe4dcac8a4bcf356b76838","family_id":"qf-str22-oscillator-model-qualification","seed":202610221024,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[2,1],\"spring\":[15,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=15/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(2, 15, Yes, Yes, Yes)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 15, Yes, Yes, Yes)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0025","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b78e71e019b9e3be2d69a0a584e0f0581af03772d854363c187f1bac4511abc","family_id":"qf-str22-oscillator-model-qualification","seed":202610221025,"semantic_template":"qf-str22-oscillator-model-qualification/{\"mass\":[3,1],\"spring\":[16,1]}"},"fields":{"title":"Positive Mass and Spring as Oscillator Model Qualifications","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=16/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. This qualification family also admits nonpositive candidates so they can be rejected. Give (normalized mass,normalized spring,mass positive?,spring positive?,both positive?).","answer":{"kind":"expression","value":"(3, 16, Yes, Yes, Yes)","explanation":"Normalize the rational signs and test both strict inequalities. Qualification alone gives no closed-form solution. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 16, Yes, Yes, Yes)."},"skills":["Positive Mass and Spring as Oscillator Model Qualifications"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0026","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"957c7478872c0918ec553621675dd97da9879bcbcd8b01bda9148aeefc2b1a59","family_id":"qf-str22-oscillator-omega-square","seed":202610221026,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[1,1],\"spring\":[2,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=2/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(1, 2, 2)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 2, 2)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0027","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"170fb8802df93451a8d069ffadc2c167891762587dca5d12b89b406b9221f2a0","family_id":"qf-str22-oscillator-omega-square","seed":202610221027,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[2,1],\"spring\":[3,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=3/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(2, 3, 3/2)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 3, 3/2)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0028","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b53c62e3f755b7c66036c6ba72291163dd8a417e361571d5c80dd6500b9fcbcf","family_id":"qf-str22-oscillator-omega-square","seed":202610221028,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[3,1],\"spring\":[4,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=4/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(3, 4, 4/3)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 4, 4/3)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0029","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11a0b07c83119cd7a727bb33c95f92eabbf24376f4c3db1d5e71ce62bd58ffef","family_id":"qf-str22-oscillator-omega-square","seed":202610221029,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[4,1],\"spring\":[5,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=5/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(4, 5, 5/4)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (4, 5, 5/4)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0030","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ada5ddae697bc9523445d559b076824a73c3fe8c2c9ed3fd0d4106244aad58dc","family_id":"qf-str22-oscillator-omega-square","seed":202610221030,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[5,1],\"spring\":[6,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=5/1 and spring coefficient k=6/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(5, 6, 6/5)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, 6, 6/5)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0031","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"668e6088ddc11bec7376c869ad41b175fda66270720940af2d580aed81257b1e","family_id":"qf-str22-oscillator-omega-square","seed":202610221031,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[1,1],\"spring\":[7,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=7/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(1, 7, 7)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 7, 7)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0032","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e7d262c0cedf3d538cded535165c146241e5e829108bd345c550cf9c46e2ceec","family_id":"qf-str22-oscillator-omega-square","seed":202610221032,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[2,1],\"spring\":[8,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=8/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(2, 8, 4)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 8, 4)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0033","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"745a33dee4f72365fe9825790410ea2b6ebfc3e3ef092df75b42eac5e0ffcb14","family_id":"qf-str22-oscillator-omega-square","seed":202610221033,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[3,1],\"spring\":[9,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=9/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(3, 9, 3)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 9, 3)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0034","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f3f806cde4365998e958b32a9f1b48c528526520b1ab514ee0c3e26f351ba21","family_id":"qf-str22-oscillator-omega-square","seed":202610221034,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[4,1],\"spring\":[10,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=10/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(4, 10, 5/2)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (4, 10, 5/2)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0035","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c502f23e5f18e8a11ba8dcc66128e3f59004f7d4e168389573817d1bf288fb3","family_id":"qf-str22-oscillator-omega-square","seed":202610221035,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[5,1],\"spring\":[11,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=5/1 and spring coefficient k=11/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(5, 11, 11/5)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, 11, 11/5)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0036","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b3619fdd36da304d493b405a03f1297b039aad93fbe2ec41e37623ce903680c5","family_id":"qf-str22-oscillator-omega-square","seed":202610221036,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[1,1],\"spring\":[12,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=12/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(1, 12, 12)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 12, 12)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0037","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e9aab231810f8ee156b4af899f4eb3310209af2767bbc1ebbb4c7181ca009935","family_id":"qf-str22-oscillator-omega-square","seed":202610221037,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[2,1],\"spring\":[13,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=13/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(2, 13, 13/2)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 13, 13/2)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0038","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1aa1deba64b283a83044c8f7f835d7a6c7a2f85a7cdffe44530cf48e92fd331b","family_id":"qf-str22-oscillator-omega-square","seed":202610221038,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[3,1],\"spring\":[14,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=14/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(3, 14, 14/3)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 14, 14/3)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0039","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ab376adcb19ce9d866d19f892f298632e478de93a5104a65dfbf07e00265430","family_id":"qf-str22-oscillator-omega-square","seed":202610221039,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[4,1],\"spring\":[15,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=15/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(4, 15, 15/4)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (4, 15, 15/4)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0040","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45ded0101a5a8548430b77714871c8c2bd5666eb150597c3847b6e7e68aaa4e8","family_id":"qf-str22-oscillator-omega-square","seed":202610221040,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[5,1],\"spring\":[16,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=5/1 and spring coefficient k=16/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(5, 16, 16/5)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, 16, 16/5)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0041","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4351b09c5c577c2992ba372893c46d7d4b151cfb57fd26a537d78c44be989bf7","family_id":"qf-str22-oscillator-omega-square","seed":202610221041,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[1,1],\"spring\":[17,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=17/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(1, 17, 17)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 17, 17)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0042","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"615a430ba8494b1d6c8e6d1fa56d7f3242ef80901de17397dbd732585d86fab8","family_id":"qf-str22-oscillator-omega-square","seed":202610221042,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[2,1],\"spring\":[18,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=18/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(2, 18, 9)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 18, 9)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0043","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"70bba81ba9bff69d65040c5cde52b1aa14e7ccbfd8db9cdbaca3cf44e1a75e3f","family_id":"qf-str22-oscillator-omega-square","seed":202610221043,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[3,1],\"spring\":[19,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=19/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(3, 19, 19/3)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 19, 19/3)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0044","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"127236e2a800a54b3eafd68ebb7c699f4e668bd9140af15e823503eb354f0e9e","family_id":"qf-str22-oscillator-omega-square","seed":202610221044,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[4,1],\"spring\":[20,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=20/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(4, 20, 5)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (4, 20, 5)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0045","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2742e79c8dbcd4c2d7e2b39eacbbab02d9e324c122437ff48bda32cca81970a2","family_id":"qf-str22-oscillator-omega-square","seed":202610221045,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[5,1],\"spring\":[21,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=5/1 and spring coefficient k=21/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(5, 21, 21/5)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, 21, 21/5)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0046","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6f7ba02065a578706c0f0164b9d158a6d9bfef1a12c1bd82ca92ecc5f249136f","family_id":"qf-str22-oscillator-omega-square","seed":202610221046,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[1,1],\"spring\":[22,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=22/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(1, 22, 22)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 22, 22)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0047","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff5d5027ff9ca5cda3513696ca14f6ecdb73fe924f772930ca0dff49bb06d206","family_id":"qf-str22-oscillator-omega-square","seed":202610221047,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[2,1],\"spring\":[23,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=23/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(2, 23, 23/2)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 23, 23/2)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0048","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f08f69a26d63065e477f32e85f55800ead2a23376fb1c0941b9489e3ae48b4e6","family_id":"qf-str22-oscillator-omega-square","seed":202610221048,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[3,1],\"spring\":[24,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=24/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(3, 24, 8)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 24, 8)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0049","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68fb9b82ae6c64fad2f3eb428590fe9ec985ec25c422b5d97ad74ed38e6ecf77","family_id":"qf-str22-oscillator-omega-square","seed":202610221049,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[4,1],\"spring\":[25,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=25/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(4, 25, 25/4)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (4, 25, 25/4)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0050","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1781dd345c2c6a19b6e011012051da1b60882783c25b07bb47d071498e3f46f","family_id":"qf-str22-oscillator-omega-square","seed":202610221050,"semantic_template":"qf-str22-oscillator-omega-square/{\"mass\":[5,1],\"spring\":[26,1]}"},"fields":{"title":"Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=5/1 and spring coefficient k=26/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Require both parameters positive. Give (mass,spring,k/m), the angular-frequency-square candidate rather than its square root.","answer":{"kind":"expression","value":"(5, 26, 26/5)","explanation":"Divide the spring coefficient by mass exactly. This ratio does not supply a sin/cos solution without the required function and derivative semantics. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (5, 26, 26/5)."},"skills":["Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0051","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e57aba969d25c94007903dc71950ae683400da2f2b1e3c980f398c8b916818f4","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221051,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-12,-2,-3],\"mass\":[1,1],\"spring\":[1,1],\"x\":-2}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-12, -2, -3], point=-2. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-12, -2, -3], -2, -20, -6, 1, 1, -26, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-12, -2, -3], -2, -20, -6, 1, 1, -26, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0052","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44ecd2288525de1fff59a5ba6d14063393425635299538bbd4d48124f861bd38","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221052,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-11,-1,-1],\"mass\":[2,1],\"spring\":[2,1],\"x\":-1}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=2/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-11, -1, -1], point=-1. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-11, -1, -1], -1, -11, -2, 2, 2, -26, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-11, -1, -1], -1, -11, -2, 2, 2, -26, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0053","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"54bff500f999f2203430a28b064a436272872552a1750bae3bf053c7ac57d3ea","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221053,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-10,0,1],\"mass\":[3,1],\"spring\":[3,1],\"x\":0}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=3/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-10, 0, 1], point=0. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-10, 0, 1], 0, -10, 2, 3, 3, -24, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-10, 0, 1], 0, -10, 2, 3, 3, -24, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0054","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"254ed60e582ac30972593523d755d9890367f3b9cfccd76651d1e847d2e05f85","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221054,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-9,1,3],\"mass\":[4,1],\"spring\":[4,1],\"x\":1}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=4/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-9, 1, 3], point=1. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-9, 1, 3], 1, -5, 6, 4, 4, 4, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-9, 1, 3], 1, -5, 6, 4, 4, 4, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0055","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0cc684eab32d843c3f5f49f6d4c7c61464ddcb22262aac784ef046b34e402d61","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221055,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-8,2,-2],\"mass\":[1,1],\"spring\":[5,1],\"x\":2}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=5/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-8, 2, -2], point=2. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-8, 2, -2], 2, -12, -4, 1, 5, -64, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-8, 2, -2], 2, -12, -4, 1, 5, -64, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0056","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb980defd4228851964d7d48522a133cc9aff3a45fd036d659b29ba960eca6b5","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221056,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-7,-2,0],\"mass\":[2,1],\"spring\":[6,1],\"x\":-2}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=6/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-7, -2, 0], point=-2. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-7, -2, 0], -2, -3, 0, 2, 6, -18, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-7, -2, 0], -2, -3, 0, 2, 6, -18, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0057","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8effb5fe88a206369ba44b6fe575356f85e0a0785a60f75050a5b25070771f8","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221057,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-6,-1,2],\"mass\":[3,1],\"spring\":[1,1],\"x\":-1}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-6, -1, 2], point=-1. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-6, -1, 2], -1, -3, 4, 3, 1, 9, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-6, -1, 2], -1, -3, 4, 3, 1, 9, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0058","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"03024a20d31647fe331a2bb5b1d7354f353abd50044c5cb01f407a156b3deadc","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221058,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-5,0,-3],\"mass\":[4,1],\"spring\":[2,1],\"x\":0}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=2/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-5, 0, -3], point=0. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-5, 0, -3], 0, -5, -6, 4, 2, -34, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-5, 0, -3], 0, -5, -6, 4, 2, -34, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0059","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"86e2cef1189616877715e4ef2242433849bfbe133820aca2a6639de097acfd17","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221059,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-4,1,-1],\"mass\":[1,1],\"spring\":[3,1],\"x\":1}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=3/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-4, 1, -1], point=1. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-4, 1, -1], 1, -4, -2, 1, 3, -14, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-4, 1, -1], 1, -4, -2, 1, 3, -14, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0060","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"895950c5b531c9474ae418f9c4f49bf10792f17034d95ebffc846cf84a4b3501","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221060,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-3,2,1],\"mass\":[2,1],\"spring\":[4,1],\"x\":2}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=4/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-3, 2, 1], point=2. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-3, 2, 1], 2, 5, 2, 2, 4, 24, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-3, 2, 1], 2, 5, 2, 2, 4, 24, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0061","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43a05ba0dd09d8f4325631a20d29aff3fa28a741b7792064fb83cc1059160520","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221061,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-2,-2,3],\"mass\":[3,1],\"spring\":[5,1],\"x\":-2}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=5/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-2, -2, 3], point=-2. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-2, -2, 3], -2, 14, 6, 3, 5, 88, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-2, -2, 3], -2, 14, 6, 3, 5, 88, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0062","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"32494ccb06ff15decabba4435cc8e22d4a15e640bca8ed3a30fc693348ffdedd","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221062,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[-1,-1,-2],\"mass\":[4,1],\"spring\":[6,1],\"x\":-1}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=6/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[-1, -1, -2], point=-1. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([-1, -1, -2], -1, -2, -4, 4, 6, -28, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([-1, -1, -2], -1, -2, -4, 4, 6, -28, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0063","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa8fc71878eebd6a32fdcc6a5e7c83ffec8dea5a6b21501b340e54e0ca93a7aa","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221063,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[0,0,0],\"mass\":[1,1],\"spring\":[1,1],\"x\":0}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[0, 0, 0], point=0. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([0, 0, 0], 0, 0, 0, 1, 1, 0, Yes)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([0, 0, 0], 0, 0, 0, 1, 1, 0, Yes)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0064","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9db9e8994d5b9e542752c034f44e26c58e706926535cc58ee18e4ec440f17641","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221064,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[1,1,2],\"mass\":[2,1],\"spring\":[2,1],\"x\":1}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=2/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[1, 1, 2], point=1. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([1, 1, 2], 1, 4, 4, 2, 2, 16, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([1, 1, 2], 1, 4, 4, 2, 2, 16, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0065","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c88d181b847740f5290e04fc47b50b79d426931b836dd5c26e3ce4e45aaf06d","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221065,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[2,2,-3],\"mass\":[3,1],\"spring\":[3,1],\"x\":2}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=3/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[2, 2, -3], point=2. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([2, 2, -3], 2, -6, -6, 3, 3, -36, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([2, 2, -3], 2, -6, -6, 3, 3, -36, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0066","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31ec1ea25a2f7a73658eb711b634c5206909119b13b058e1ee0ac14dff2bfbbc","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221066,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[3,-2,-1],\"mass\":[4,1],\"spring\":[4,1],\"x\":-2}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=4/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[3, -2, -1], point=-2. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([3, -2, -1], -2, 3, -2, 4, 4, 4, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([3, -2, -1], -2, 3, -2, 4, 4, 4, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0067","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b5e5430194441d7a65110690639b8b4070545961846c3fca00681640e73012d","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221067,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[4,-1,1],\"mass\":[1,1],\"spring\":[5,1],\"x\":-1}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=5/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[4, -1, 1], point=-1. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([4, -1, 1], -1, 6, 2, 1, 5, 32, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([4, -1, 1], -1, 6, 2, 1, 5, 32, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0068","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bef2106903ffde4d1b30e9a57139cfcb013d3cd681a5dbdc11149d820c93ae10","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221068,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[5,0,3],\"mass\":[2,1],\"spring\":[6,1],\"x\":0}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=6/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[5, 0, 3], point=0. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([5, 0, 3], 0, 5, 6, 2, 6, 42, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([5, 0, 3], 0, 5, 6, 2, 6, 42, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0069","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"741e9c274a4044d1c47052df5758116b605e0d45eb2844577539c1842e15e21d","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221069,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[6,1,-2],\"mass\":[3,1],\"spring\":[1,1],\"x\":1}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[6, 1, -2], point=1. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([6, 1, -2], 1, 5, -4, 3, 1, -7, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([6, 1, -2], 1, 5, -4, 3, 1, -7, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0070","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b8cd9ede089299fdd539f06918cdee68a0c0e888b4ebac2353ebea833928fa4","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221070,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[7,2,0],\"mass\":[4,1],\"spring\":[2,1],\"x\":2}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=2/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[7, 2, 0], point=2. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([7, 2, 0], 2, 11, 0, 4, 2, 22, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([7, 2, 0], 2, 11, 0, 4, 2, 22, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0071","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d6dc60bf595f1b93abf2b5eca51ba7c17425d06214214f974680646b0c6e374f","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221071,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[8,-2,2],\"mass\":[1,1],\"spring\":[3,1],\"x\":-2}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=3/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[8, -2, 2], point=-2. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([8, -2, 2], -2, 20, 4, 1, 3, 64, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([8, -2, 2], -2, 20, 4, 1, 3, 64, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0072","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"08b6bb64ce656e52a6fbd1b3900d8c19834e563c78fe695bbed7e4a454fb464a","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221072,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[9,-1,-3],\"mass\":[2,1],\"spring\":[4,1],\"x\":-1}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=4/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[9, -1, -3], point=-1. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([9, -1, -3], -1, 7, -6, 2, 4, 16, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([9, -1, -3], -1, 7, -6, 2, 4, 16, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0073","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b2e6df537031270c4be6a2d046e2ac9fd932c21d461855f6db5e5786cab61cf","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221073,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[10,0,-1],\"mass\":[3,1],\"spring\":[5,1],\"x\":0}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=5/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[10, 0, -1], point=0. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([10, 0, -1], 0, 10, -2, 3, 5, 44, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([10, 0, -1], 0, 10, -2, 3, 5, 44, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0074","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ffec4497b6b40350febf2d68044b033215d1b416a9a90bc8cd8a99dbbd5601fb","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221074,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[11,1,1],\"mass\":[4,1],\"spring\":[6,1],\"x\":1}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=6/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[11, 1, 1], point=1. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([11, 1, 1], 1, 13, 2, 4, 6, 86, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([11, 1, 1], 1, 13, 2, 4, 6, 86, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0075","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6153b522d6b96e74cbdceeb118c8a938912d4d127c4692d5b8b43af697017129","family_id":"qf-str22-oscillator-polynomial-residual","seed":202610221075,"semantic_template":"qf-str22-oscillator-polynomial-residual/{\"coefficients\":[12,2,3],\"mass\":[1,1],\"spring\":[1,1],\"x\":2}"},"fields":{"title":"A Polynomial Oscillator Residual at One Specified Point","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. Ascending-power coefficients=[12, 2, 3], point=2. Give (coefficients,point,x(point),x''(point),mass,spring,m*x''+k*x,zero at this point?).","answer":{"kind":"expression","value":"([12, 2, 3], 2, 28, 6, 1, 1, 34, No)","explanation":"Differentiate twice and recompute the signed residual. A zero at one point is not an interval-wide solution or a stability proof. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is ([12, 2, 3], 2, 28, 6, 1, 1, 34, No)."},"skills":["A Polynomial Oscillator Residual at One Specified Point"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0076","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff4f0cebcf4283be95b8a4410028f383bdf4a2ed6762d25a0900bf6d70453369","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221076,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[1,1],\"position\":[-10,1],\"spring\":[1,1],\"velocity\":[-3,1]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=-10/1, velocity=-3/1. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(1, 1, -10, -3, 9/2, 50, 109/2)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 1, -10, -3, 9/2, 50, 109/2)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0077","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b734213a235c085c4125d9193abab0b5f500e2c74bbbb8851a25377fb9dc462","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221077,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[2,1],\"position\":[-9,2],\"spring\":[2,1],\"velocity\":[-2,2]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=2/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=-9/2, velocity=-2/2. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(2, 2, -9/2, -1, 1, 81/4, 85/4)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 2, -9/2, -1, 1, 81/4, 85/4)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0078","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dee6d2e034f047c69d24bd747129bb33cfc1fcd52da53f2a0d227b4370012bfa","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221078,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[3,1],\"position\":[-8,3],\"spring\":[3,1],\"velocity\":[-1,3]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=3/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=-8/3, velocity=-1/3. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(3, 3, -8/3, -1/3, 1/6, 32/3, 65/6)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 3, -8/3, -1/3, 1/6, 32/3, 65/6)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0079","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49ece595dc90354d342ee7eaace380639a16693cfe1a38067248437197356472","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221079,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[4,1],\"position\":[-7,1],\"spring\":[4,1],\"velocity\":[0,4]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=4/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=-7/1, velocity=0/4. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(4, 4, -7, 0, 0, 98, 98)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (4, 4, -7, 0, 0, 98, 98)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0080","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1510bc9fe8c9151c85e9425b4115d346027f5ccd55142dd120d8635a6bc95747","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221080,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[1,1],\"position\":[-6,2],\"spring\":[5,1],\"velocity\":[1,1]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=5/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=-6/2, velocity=1/1. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(1, 5, -3, 1, 1/2, 45/2, 23)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 5, -3, 1, 1/2, 45/2, 23)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0081","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"044aca8aff951a4fc6bc5beeea252073f7c544c1ba1ded857e5d9a0a1d3cf132","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221081,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[2,1],\"position\":[-5,3],\"spring\":[6,1],\"velocity\":[2,2]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=6/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=-5/3, velocity=2/2. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(2, 6, -5/3, 1, 1, 25/3, 28/3)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 6, -5/3, 1, 1, 25/3, 28/3)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0082","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82c16c2c7a9b6626c8f5896b0ac8cc280b052b762af8b00485e9b529bf4e27d6","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221082,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[3,1],\"position\":[-4,1],\"spring\":[1,1],\"velocity\":[3,3]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=-4/1, velocity=3/3. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(3, 1, -4, 1, 3/2, 8, 19/2)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 1, -4, 1, 3/2, 8, 19/2)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0083","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b5ba8103afb0af6deac1ff36da1102d1280d18672d28914c147bc95f41d40c8","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221083,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[4,1],\"position\":[-3,2],\"spring\":[2,1],\"velocity\":[-3,4]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=2/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=-3/2, velocity=-3/4. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(4, 2, -3/2, -3/4, 9/8, 9/4, 27/8)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (4, 2, -3/2, -3/4, 9/8, 9/4, 27/8)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0084","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6968c9602d3c4a2ce277784c9241d653e93f9c7c5ab58cafec00de511f6e01af","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221084,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[1,1],\"position\":[-2,3],\"spring\":[3,1],\"velocity\":[-2,1]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=3/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=-2/3, velocity=-2/1. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(1, 3, -2/3, -2, 2, 2/3, 8/3)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 3, -2/3, -2, 2, 2/3, 8/3)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0085","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2656ffa74133456f4c4736b305a799168b33b1e2735d04fcf5912cdff48f5e7f","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221085,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[2,1],\"position\":[-1,1],\"spring\":[4,1],\"velocity\":[-1,2]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=4/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=-1/1, velocity=-1/2. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(2, 4, -1, -1/2, 1/4, 2, 9/4)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 4, -1, -1/2, 1/4, 2, 9/4)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0086","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5f7b36252b79a361f15b2877c6230620569d552204cb743bd02c4c7afd56c2b7","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221086,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[3,1],\"position\":[0,2],\"spring\":[5,1],\"velocity\":[0,3]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=5/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=0/2, velocity=0/3. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(3, 5, 0, 0, 0, 0, 0)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 5, 0, 0, 0, 0, 0)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0087","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"86adb8e524610c127beee671a8662171663ee6ea93d6f702a416f2e5f2654708","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221087,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[4,1],\"position\":[1,3],\"spring\":[6,1],\"velocity\":[1,4]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=6/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=1/3, velocity=1/4. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(4, 6, 1/3, 1/4, 1/8, 1/3, 11/24)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (4, 6, 1/3, 1/4, 1/8, 1/3, 11/24)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0088","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"707edc3185d42b42e051eb9a358e31aa225363a9b62cc41341ebd817f5f3fbd6","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221088,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[1,1],\"position\":[2,1],\"spring\":[1,1],\"velocity\":[2,1]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=2/1, velocity=2/1. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(1, 1, 2, 2, 2, 2, 4)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 1, 2, 2, 2, 2, 4)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0089","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6dfacdb011619ec056c705effe77b97f2c65882d4065da67a669b8e43335c109","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221089,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[2,1],\"position\":[3,2],\"spring\":[2,1],\"velocity\":[3,2]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=2/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=3/2, velocity=3/2. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(2, 2, 3/2, 3/2, 9/4, 9/4, 9/2)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 2, 3/2, 3/2, 9/4, 9/4, 9/2)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0090","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6f35cb4ee992f0c3f498f8e3345197b8e853b37e336df89f9302495f1dacd0c","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221090,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[3,1],\"position\":[4,3],\"spring\":[3,1],\"velocity\":[-3,3]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=3/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=4/3, velocity=-3/3. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(3, 3, 4/3, -1, 3/2, 8/3, 25/6)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 3, 4/3, -1, 3/2, 8/3, 25/6)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0091","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b4b5a5bd9d3c4ad3aadcbaacc1e3c9a74e9d466472b2199f0dd699d80e9ea7e4","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221091,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[4,1],\"position\":[5,1],\"spring\":[4,1],\"velocity\":[-2,4]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=4/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=5/1, velocity=-2/4. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(4, 4, 5, -1/2, 1/2, 50, 101/2)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (4, 4, 5, -1/2, 1/2, 50, 101/2)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0092","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e7805233a502cfe2aa145280aa6d611403d6a1f3f091fddaa221b2ec797ce0fd","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221092,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[1,1],\"position\":[6,2],\"spring\":[5,1],\"velocity\":[-1,1]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=5/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=6/2, velocity=-1/1. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(1, 5, 3, -1, 1/2, 45/2, 23)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 5, 3, -1, 1/2, 45/2, 23)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0093","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d093a3570d178869931782b86bcf0fa7200c25aad169ed85977ed394e48452e","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221093,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[2,1],\"position\":[7,3],\"spring\":[6,1],\"velocity\":[0,2]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=6/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=7/3, velocity=0/2. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(2, 6, 7/3, 0, 0, 49/3, 49/3)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 6, 7/3, 0, 0, 49/3, 49/3)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0094","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"59d7b0814d54eb8d34f16128be7ab31b632a5cbec3346006be6beb5bf419b1ce","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221094,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[3,1],\"position\":[8,1],\"spring\":[1,1],\"velocity\":[1,3]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=8/1, velocity=1/3. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(3, 1, 8, 1/3, 1/6, 32, 193/6)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 1, 8, 1/3, 1/6, 32, 193/6)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0095","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"efa72a76bdc59bdabfea3ec55eb812911fc51f746d2138a5ca95799b3c29f18f","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221095,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[4,1],\"position\":[9,2],\"spring\":[2,1],\"velocity\":[2,4]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=2/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=9/2, velocity=2/4. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(4, 2, 9/2, 1/2, 1/2, 81/4, 83/4)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (4, 2, 9/2, 1/2, 1/2, 81/4, 83/4)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0096","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7f115d2fd24341eedffd66baf3511279a04ce104b33005a8d96d9d1b725c5a0","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221096,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[1,1],\"position\":[10,3],\"spring\":[3,1],\"velocity\":[3,1]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=3/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=10/3, velocity=3/1. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(1, 3, 10/3, 3, 9/2, 50/3, 127/6)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 3, 10/3, 3, 9/2, 50/3, 127/6)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0097","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb8ca05ed40cc1e78d2c67bf932994eb154efe1bc9a0084cc900ae75d37c5e55","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221097,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[2,1],\"position\":[11,1],\"spring\":[4,1],\"velocity\":[-3,2]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=2/1 and spring coefficient k=4/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=11/1, velocity=-3/2. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(2, 4, 11, -3/2, 9/4, 242, 977/4)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (2, 4, 11, -3/2, 9/4, 242, 977/4)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0098","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb350cca2de9690ee79a6670ffca9aa057e8e91f991a1e8a4a46ed59abc3e4c5","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221098,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[3,1],\"position\":[12,2],\"spring\":[5,1],\"velocity\":[-2,3]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=3/1 and spring coefficient k=5/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=12/2, velocity=-2/3. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(3, 5, 6, -2/3, 2/3, 90, 272/3)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (3, 5, 6, -2/3, 2/3, 90, 272/3)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0099","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1bd284813141635e6b51c67dc9830f39a7a24304994adfa8858b65ffa75b97d8","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221099,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[4,1],\"position\":[13,3],\"spring\":[6,1],\"velocity\":[-1,4]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=4/1 and spring coefficient k=6/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=13/3, velocity=-1/4. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(4, 6, 13/3, -1/4, 1/8, 169/3, 1355/24)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (4, 6, 13/3, -1/4, 1/8, 169/3, 1355/24)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261022-r01-harmonic-oscillator-model-0100","batch_id":"batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd75514da2696ba63f88d8b1a4ad96dd186a13adf3c3d71034d051ec6d30211a","family_id":"qf-str22-oscillator-energy-snapshot","seed":202610221100,"semantic_template":"qf-str22-oscillator-energy-snapshot/{\"mass\":[1,1],\"position\":[14,1],\"spring\":[1,1],\"velocity\":[0,1]}"},"fields":{"title":"Kinetic and Potential Energy at One State Only","stem":"Use the undamped linear oscillator model m*x''+k*x=0, with mass m=1/1 and spring coefficient k=1/1. Use exact rationals, normalize denominator signs, reduce fractions and omit /1 for integer outputs. Positive m and k qualify this model; this slice establishes no sin/cos closed-form solution or stability theorem. position=14/1, velocity=0/1. Give (mass,spring,position,velocity,m*v²/2,k*x²/2,total energy). Signed position and velocity are squared, not clipped.","answer":{"kind":"expression","value":"(1, 1, 14, 0, 0, 98, 98)","explanation":"Compute the exact two quadratic terms. A single-state energy value does not prove conservation along a trajectory. This explicit exact slice is not a general theorem or empirical assertion. The exact result, retaining the requested order, is (1, 1, 14, 0, 0, 98, 98)."},"skills":["Kinetic and Potential Energy at One State Only"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}