Positive Oscillator Parameters, Point Residuals and State-Energy Values
This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.
This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.
batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations · 2026-10-13 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
1. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(-1, -8, No, No, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
2. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(0, -7, No, No, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
3. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(1, -6, Yes, No, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
4. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(2, -5, Yes, No, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
5. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(3, -4, Yes, No, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
6. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(-1, -3, No, No, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
7. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(0, -2, No, No, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
8. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(1, -1, Yes, No, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
9. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(2, 0, Yes, No, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
10. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(3, 1, Yes, Yes, Yes)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
11. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(-1, 2, No, Yes, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
12. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(0, 3, No, Yes, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
13. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(1, 4, Yes, Yes, Yes)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
14. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(2, 5, Yes, Yes, Yes)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
15. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(3, 6, Yes, Yes, Yes)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
16. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(-1, 7, No, Yes, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
17. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(0, 8, No, Yes, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
18. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(1, 9, Yes, Yes, Yes)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
19. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(2, 10, Yes, Yes, Yes)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
20. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(3, 11, Yes, Yes, Yes)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
21. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(-1, 12, No, Yes, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
22. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(0, 13, No, Yes, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
23. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(1, 14, Yes, Yes, Yes)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
24. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(2, 15, Yes, Yes, Yes)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
25. Positive Mass and Spring as Oscillator Model Qualifications
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?).
View answer and explanation
(3, 16, Yes, Yes, Yes)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-model-qualification · candidate
26. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(1, 2, 2)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
27. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(2, 3, 3/2)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
28. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(3, 4, 4/3)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
29. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(4, 5, 5/4)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
30. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(5, 6, 6/5)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
31. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(1, 7, 7)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
32. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(2, 8, 4)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
33. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(3, 9, 3)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
34. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(4, 10, 5/2)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
35. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(5, 11, 11/5)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
36. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(1, 12, 12)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
37. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(2, 13, 13/2)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
38. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(3, 14, 14/3)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
39. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(4, 15, 15/4)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
40. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(5, 16, 16/5)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
41. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(1, 17, 17)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
42. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(2, 18, 9)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
43. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(3, 19, 19/3)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
44. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(4, 20, 5)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
45. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(5, 21, 21/5)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
46. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(1, 22, 22)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
47. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(2, 23, 23/2)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
48. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(3, 24, 8)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
49. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(4, 25, 25/4)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
50. Compute Angular Frequency Squared, Not a Claimed Closed-Form Solution
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.
View answer and explanation
(5, 26, 26/5)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-omega-square · candidate
51. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-12, -2, -3], -2, -20, -6, 1, 1, -26, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
52. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-11, -1, -1], -1, -11, -2, 2, 2, -26, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
53. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-10, 0, 1], 0, -10, 2, 3, 3, -24, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
54. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-9, 1, 3], 1, -5, 6, 4, 4, 4, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
55. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-8, 2, -2], 2, -12, -4, 1, 5, -64, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
56. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-7, -2, 0], -2, -3, 0, 2, 6, -18, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
57. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-6, -1, 2], -1, -3, 4, 3, 1, 9, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
58. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-5, 0, -3], 0, -5, -6, 4, 2, -34, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
59. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-4, 1, -1], 1, -4, -2, 1, 3, -14, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
60. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-3, 2, 1], 2, 5, 2, 2, 4, 24, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
61. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-2, -2, 3], -2, 14, 6, 3, 5, 88, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
62. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([-1, -1, -2], -1, -2, -4, 4, 6, -28, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
63. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([0, 0, 0], 0, 0, 0, 1, 1, 0, Yes)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
64. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([1, 1, 2], 1, 4, 4, 2, 2, 16, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
65. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([2, 2, -3], 2, -6, -6, 3, 3, -36, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
66. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([3, -2, -1], -2, 3, -2, 4, 4, 4, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
67. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([4, -1, 1], -1, 6, 2, 1, 5, 32, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
68. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([5, 0, 3], 0, 5, 6, 2, 6, 42, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
69. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([6, 1, -2], 1, 5, -4, 3, 1, -7, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
70. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([7, 2, 0], 2, 11, 0, 4, 2, 22, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
71. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([8, -2, 2], -2, 20, 4, 1, 3, 64, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
72. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([9, -1, -3], -1, 7, -6, 2, 4, 16, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
73. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([10, 0, -1], 0, 10, -2, 3, 5, 44, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
74. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([11, 1, 1], 1, 13, 2, 4, 6, 86, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
75. A Polynomial Oscillator Residual at One Specified Point
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?).
View answer and explanation
([12, 2, 3], 2, 28, 6, 1, 1, 34, No)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-polynomial-residual · candidate
76. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(1, 1, -10, -3, 9/2, 50, 109/2)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
77. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(2, 2, -9/2, -1, 1, 81/4, 85/4)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
78. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(3, 3, -8/3, -1/3, 1/6, 32/3, 65/6)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
79. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(4, 4, -7, 0, 0, 98, 98)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
80. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(1, 5, -3, 1, 1/2, 45/2, 23)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
81. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(2, 6, -5/3, 1, 1, 25/3, 28/3)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
82. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(3, 1, -4, 1, 3/2, 8, 19/2)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
83. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(4, 2, -3/2, -3/4, 9/8, 9/4, 27/8)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
84. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(1, 3, -2/3, -2, 2, 2/3, 8/3)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
85. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(2, 4, -1, -1/2, 1/4, 2, 9/4)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
86. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(3, 5, 0, 0, 0, 0, 0)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
87. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(4, 6, 1/3, 1/4, 1/8, 1/3, 11/24)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
88. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(1, 1, 2, 2, 2, 2, 4)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
89. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(2, 2, 3/2, 3/2, 9/4, 9/4, 9/2)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
90. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(3, 3, 4/3, -1, 3/2, 8/3, 25/6)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
91. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(4, 4, 5, -1/2, 1/2, 50, 101/2)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
92. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(1, 5, 3, -1, 1/2, 45/2, 23)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
93. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(2, 6, 7/3, 0, 0, 49/3, 49/3)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
94. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(3, 1, 8, 1/3, 1/6, 32, 193/6)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
95. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(4, 2, 9/2, 1/2, 1/2, 81/4, 83/4)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
96. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(1, 3, 10/3, 3, 9/2, 50/3, 127/6)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
97. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(2, 4, 11, -3/2, 9/4, 242, 977/4)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
98. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(3, 5, 6, -2/3, 2/3, 90, 272/3)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
99. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(4, 6, 13/3, -1/4, 1/8, 169/3, 1355/24)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
100. Kinetic and Potential Energy at One State Only
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.
View answer and explanation
(1, 1, 14, 0, 0, 98, 98)
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).
Reviewed step-by-step derivation
Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate
Canonical question JSONL · English JSONL · Validation report