Spring-mass systems
Problem 6.727 · hard
A weight of \( \displaystyle 32 \) lb stretches a spring \( \displaystyle 2 \) ft. A damping force of \( \displaystyle 4 \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{4} \) ft below equilibrium with velocity \( \displaystyle -2 \) ft/s (down is positive). Find the position x(t).
- \[ \frac{32}{2} = 16 \]Hooke's law: k = weight / stretch (lb/ft).✓ Proved
- \[ \frac{32}{32} = 1 \]m = weight / g, with g = 32 ft/s².✓ Proved
- The motion satisfies m x″ + c x′ + k x = 0: 1x″ + 4x′ + 16x = 0 (x measured downward from equilibrium).Reviewed
- \[ 4 \frac{d}{d t} \frac{e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{3} \right)}}{2} + \frac{d^{2}}{d t^{2}} \frac{e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{3} \right)}}{2} + 8 e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{3} \right)} = 0 \]This x(t) satisfies it ...✓ Proved
- \[ \left. \frac{e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{3} \right)}}{2} \right|_{\substack{ t=0 }} = \frac{1}{4} \]... starts in the right place ...✓ Proved
- \[ \left. \frac{d}{d t} \frac{e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{3} \right)}}{2} \right|_{\substack{ t=0 }} = -2 \]... with the right velocity.✓ Proved
Answer \( x(t) = \frac{e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{3} \right)}}{2} \)
Lines: 5 proved, 1 reviewed. 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 6 | ✓ 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: pass — The solution correctly derives the parameters, sets up the differential equation with consistent sign conventions, and verifies that the stated answer satisfies both the ODE and the initial conditions.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-20 — The solution correctly derives the parameters, sets up the differential equation with consistent sign conventions, and verifies that the stated answer satisfies both the ODE and the initial conditions.gpt-oss:20b: pass 2026-10-20qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly derives the parameters, sets up the differential equation with the correct sign convention, and verifies that the proposed solution satisfies both the equation and the initial conditions.gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly derives the parameters and verifies that the stated answer satisfies the differential equation and initial conditions.gpt-oss:20b: fail (error) 2026-10-19 — The characteristic equation for m=1, c=4, k=16 is r^2+4r+16=0, giving roots -2±4√3 i. Thus the general solution should involve e^{-2t}[C1 cos(4√3 t)+C2 sin(4√3 t)]. The proposed solution uses 2√3 instead of 4√3, so it does not satisfy the differential equation and cannot be correct.
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 2026-10-20 with SymPy 1.14.0.