Spring-mass systems
Problem 6.634 · hard
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 50 \) N/m. A damping force of \( \displaystyle 10 \) times the velocity acts on it. It is released from rest \( \displaystyle \frac{1}{2} \) m above equilibrium. Find the position x(t).
- The motion satisfies m x″ + c x′ + k x = 0: 2x″ + 10x′ + 50x = 0 (x measured downward from equilibrium).
- \[ 10 \frac{d}{d t} \left(- \frac{\sqrt{3} e^{- \frac{5 t}{2}} \sin{\left(\frac{5 \sqrt{3} t}{2} + \frac{\pi}{3} \right)}}{3}\right) + 2 \frac{d^{2}}{d t^{2}} \left(- \frac{\sqrt{3} e^{- \frac{5 t}{2}} \sin{\left(\frac{5 \sqrt{3} t}{2} + \frac{\pi}{3} \right)}}{3}\right) - \frac{50 \sqrt{3} e^{- \frac{5 t}{2}} \sin{\left(\frac{5 \sqrt{3} t}{2} + \frac{\pi}{3} \right)}}{3} = 0 \]This x(t) satisfies it ...✓ Proved
- \[ \left. \frac{- \sqrt{3} e^{- \frac{5 t}{2}} \sin{\left(\frac{5 \sqrt{3} t}{2} + \frac{\pi}{3} \right)}}{3} \right|_{\substack{ t=0 }} = - \frac{1}{2} \]... starts in the right place ...✓ Proved
- \[ \left. \frac{d}{d t} \frac{- \sqrt{3} e^{- \frac{5 t}{2}} \sin{\left(\frac{5 \sqrt{3} t}{2} + \frac{\pi}{3} \right)}}{3} \right|_{\substack{ t=0 }} = 0 \]... with the right velocity.✓ Proved
Answer \( x(t) = - \frac{\sqrt{3} e^{- \frac{5 t}{2}} \sin{\left(\frac{5 \sqrt{3} t}{2} + \frac{\pi}{3} \right)}}{3} \)
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: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The initial condition is inconsistent with the coordinate system. The problem states the mass is released 1/2 m above equilibrium, but the solution d
Every verdict on record (6)
qwen3.6:27b-mlx: inconclusive 2026-10-18 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The initial condition is inconsistent with the coordinate system. The problem states the mass is released 1/2 m above equilibrium, but the solution dgpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: inconclusive 2026-10-16 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The initial condition is inconsistent with the coordinate system. The problem states the mass is released 1/2 m *above* equilibrium, but the solutiongpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-15 — The initial condition is stated as '1/2 m above equilibrium', but with x measured downward, this corresponds to x(0) = -1/2. The solution checks x(0) = -1/2, which is consistent with the setup. However, the stated answer in the problem description is x(t) = - \frac{\sqrt{3} e^{- \frac{5 t}{2}} \sin{\left(\frac{5 \sqrt{3} t}{2} + \frac{\pi}{3} \right)}}{3}. Let's check the sign. At t=0, sin(pi/3) = sqrt(3)/2. So x(0) = - (sqrt(3)/3) * (sqrt(3)/2) = - (3/6) = -1/2. This matches. The derivative at t=0: x'(t) involves cos and sin terms. The check says x'(0)=0. Let's verify. x(t) = A e^{-bt} sin(wt + phi). x'(t) = A e^{-bt} (-b sin(wt+phi) + w cos(wt+phi)). At t=0: x'(0) = A (-b sin(phi) + w cos(phi)). Here A = -sqrt(3)/3, b = 5/2, w = 5sqrt(3)/2, phi = pi/3. sin(pi/3)=sqrt(3)/2, cos(pi/3)=1/2. x'(0) = (-sqrt(3)/3) * (-(5/2)(sqrt(3)/2) + (5sqrt(3)/2)(1/2)) = (-sqrt(3)/3) * (-5sqrt(3)/4 + 5sqrt(3)/4) = 0. So the math is correct. Why did I think it might fail? The prompt asks to review the solution. The solution consists of a sentence setting up the ODE and three equation checks. The sentence says 'x measured downward'. The problem says 'released from rest 1/2 m above equilibrium'. If x is downward, above equilibrium is negative x. So x(0) = -1/2. The check confirms x(0) = -1/2. The check confirms x'(0) = 0. The check confirms the ODE is satisfied. The solution is correct. Wait, is there a subtle error? 'A damping force of 10 times the velocity acts on it.' Usually damping force opposes motion, so F_d = -cv. The equation is mx'' = -kx - cv. So mx'' + cv + kx = 0. This is standard. The setup is correct. The checks are correct. The verdict should be pass.gpt-oss:20b: pass 2026-10-15
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 2026-10-18 with SymPy 1.14.0.