Spring-mass systems
Problem 6.659 · hard
A \( \displaystyle 1 \) kg mass hangs from a spring with constant \( \displaystyle 9 \) N/m. A damping force of \( \displaystyle \frac{15}{2} \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{2} \) m above equilibrium with velocity \( \displaystyle 1 \) m/s (down is positive). Find the position x(t).
- The motion satisfies m x″ + c x′ + k x = 0: 1x″ + 15/2x′ + 9x = 0 (x measured downward from equilibrium).
- \[ \frac{15 \frac{d}{d t} \left(- \frac{e^{- 6 t}}{18} - \frac{4 e^{- \frac{3 t}{2}}}{9}\right)}{2} + \frac{d^{2}}{d t^{2}} \left(- \frac{e^{- 6 t}}{18} - \frac{4 e^{- \frac{3 t}{2}}}{9}\right) - \frac{e^{- 6 t}}{2} - 4 e^{- \frac{3 t}{2}} = 0 \]This x(t) satisfies it ...✓ Proved
- \[ \left. \frac{\left(-1\right) e^{- 6 t}}{18} - \frac{4 e^{- \frac{3 t}{2}}}{9} \right|_{\substack{ t=0 }} = - \frac{1}{2} \]... starts in the right place ...✓ Proved
- \[ \left. \frac{d}{d t} \left(\frac{\left(-1\right) e^{- 6 t}}{18} - \frac{4 e^{- \frac{3 t}{2}}}{9}\right) \right|_{\substack{ t=0 }} = 1 \]... with the right velocity.✓ Proved
Answer \( x(t) = - \frac{e^{- 6 t}}{18} - \frac{4 e^{- \frac{3 t}{2}}}{9} \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | mpmath's ODE solver integrates the spring from t = 0 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution fails to derive the position function from the differential equation and initial conditions. It merely verifies that the stated answer satisfies the ODE and initial conditions, which does not constitute a solution process (e.g., finding roots of the characteristic equation and solving for constants).
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution fails to derive the position function from the differential equation and initial conditions. It merely verifies that the stated answer satisfies the ODE and initial conditions, which does not constitute a solution process (e.g., finding roots of the characteristic equation and solving for constants).gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: fail (error) 2026-10-16 — The initial position is stated as 1/2 m above equilibrium, which corresponds to x(0) = -1/2 given the 'down is positive' convention. However, the provided solution x(t) yields x(0) = -1/2, which is correct for the position, but the problem states the mass starts 1/2 m *above* equilibrium. Wait, if down is positive, above is negative. So x(0) = -1/2 is correct. Let's re-read carefully. 'Starts 1/2 m above equilibrium'. Down is positive. So x(0) = -0.5. The solution gives x(0) = -1/18 - 4/9 = -1/18 - 8/18 = -9/18 = -1/2. This matches. Velocity is 1 m/s (down is positive). x'(0) = 1. The solution satisfies the ODE and ICs. Why would it fail? Let's check the ODE setup. m=1, c=15/2, k=9. Characteristic equation: r^2 + 7.5r + 9 = 0. Roots: r = (-7.5 +/- sqrt(56.25 - 36))/2 = (-7.5 +/- sqrt(20.25))/2 = (-7.5 +/- 4.5)/2. r1 = -3/2 = -1.5, r2 = -12/2 = -6. General solution: x(t) = c1 e^{-1.5t} + c2 e^{-6t}. x(0) = c1 + c2 = -1/2. x'(t) = -1.5 c1 e^{-1.5t} - 6 c2 e^{-6t}. x'(0) = -1.5 c1 - 6 c2 = 1. Multiply first eq by 1.5: 1.5 c1 + 1.5 c2 = -0.75. Add to second: -4.5 c2 = 0.25 => c2 = -1/18. Then c1 = -1/2 - (-1/18) = -9/18 + 1/18 = -8/18 = -4/9. So x(t) = -4/9 e^{-1.5t} - 1/18 e^{-6t}. This matches the stated answer. The solution is correct. Wait, did I miss something? The prompt asks me to find what is wrong. If it's correct, I should pass. But usually these adversarial reviews have a trick. Let's look at the sentence: 'The motion satisfies m x″ + c x′ + k x = 0'. This is standard. 'x measured downward from equilibrium'. Standard. The checks are algebraic verifications. Is there a sign error in the problem statement interpretation? 'Damping force of 15/2 times the velocity acts on it'. Damping opposes motion, so force is -cv. Equation is mx'' = -kx - cv. So mx'' + cv + kx = 0. Correct. 'Starts 1/2 m above equilibrium'. x(0) = -1/2. Correct. 'Velocity 1 m/s (down is positive)'. x'(0) = 1. Correct. The solution seems perfectly sound. Why would it be a fail? Perhaps the 'Stated answer' in the problem description is just a label, and the solution derives it? No, the solution just verifies it. Is the verification sufficient? It checks the ODE and ICs. By uniqueness theorem, this is sufficient. I will pass it.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The differential equation is correctly set up with the given parameters and sign conventions. The proposed solution satisfies the ODE and the initial conditions as verified by the algebraic checks.gpt-oss:20b: fail (error) 2026-10-15 — The derivative at t=0 equals 5/3, not the required initial velocity of 1 m/s, so the proposed solution does not satisfy the initial conditions.
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/spring_mass, checked 2026-10-19 with SymPy 1.14.0.