# 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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0001/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0002/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0003/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0004/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0005/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0006/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0007/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0008/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0009/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0010/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0011/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0012/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0013/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0014/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0015/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0016/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0017/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0018/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0019/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0020/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0021/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0022/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0023/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0024/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0025/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0026/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0027/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0028/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0029/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0030/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0031/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0032/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0033/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0034/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0035/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0036/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0037/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0038/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0039/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0040/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0041/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0042/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0043/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0044/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0045/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0046/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0047/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0048/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0049/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0050/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0051/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0052/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0053/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0054/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0055/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0056/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0057/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0058/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0059/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0060/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0061/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0062/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0063/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0064/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0065/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0066/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0067/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0068/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0069/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0070/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0071/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0072/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0073/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0074/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0075/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0076/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0077/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0078/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0079/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0080/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0081/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0082/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0083/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0084/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0085/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0086/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0087/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0088/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0089/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0090/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0091/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0092/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0093/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0094/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0095/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0096/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0097/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0098/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0099/)

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](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261022-r01-harmonic-oscillator-model-0100/)

Linear harmonic-oscillator model · qf-str22-oscillator-energy-snapshot · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261013-advance-20261022-r01-harmonic-oscillator-model-foundations/validation-report.json)