Spring-mass systems
Problem 6.637 · medium
A weight of \( \displaystyle 64 \) lb stretches a spring \( \displaystyle \frac{1}{3} \) ft. There is no damping. It is released from rest \( \displaystyle \frac{1}{2} \) ft above equilibrium. Find the position x(t).
- \[ \frac{64}{1 \cdot \frac{1}{3}} = 192 \]Hooke's law: k = weight / stretch (lb/ft).✓ Proved
- \[ \frac{64}{32} = 2 \]m = weight / g, with g = 32 ft/s².✓ Proved
- The motion satisfies m x″ + c x′ + k x = 0: 2x″ + 0x′ + 192x = 0 (x measured downward from equilibrium).
- \[ - 96 \cos{\left(4 \sqrt{6} t \right)} + 2 \frac{d^{2}}{d t^{2}} \left(- \frac{\cos{\left(4 \sqrt{6} t \right)}}{2}\right) = 0 \]This x(t) satisfies it ...✓ Proved
- \[ \left. \frac{\left(-1\right) \cos{\left(4 \sqrt{6} t \right)}}{2} \right|_{\substack{ t=0 }} = - \frac{1}{2} \]... starts in the right place ...✓ Proved
- \[ \left. \frac{d}{d t} \frac{\left(-1\right) \cos{\left(4 \sqrt{6} t \right)}}{2} \right|_{\substack{ t=0 }} = 0 \]... with the right velocity.✓ Proved
Answer \( x(t) = - \frac{\cos{\left(4 \sqrt{6} t \right)}}{2} \)
Lines: 5 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | Not checked | — | 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: fail (error) — The sign convention is inconsistent. The problem states the mass is released 1/2 ft *above* equilibrium, which corresponds to an initial position of x(0) = -1/2 if x is measured downward. However, the stated answer x(t) = -cos(4√6 t)/2 yields x(0) = -1/2, which is correct for the downward convention. But wait, let's re-read carefully. If x is measured downward, 'above equilibrium' is negative. So x(0) = -1/2. The solution checks out algebraically. Let me double check the physics. k = 192, m = 2. omega = sqrt(192/2) = sqrt(96) = 4*sqrt(6). Correct. x(t) = A cos(omega t) + B sin(omega t). x(0) = -1/2 => A = -1/2. x'(0) = 0 => B = 0. So x(t) = -1/2 cos(4*sqrt(6) t). The solution is actually correct. Why did I think it was wrong? Ah, I need to check if the sentence in step 3 defines the direction correctly. 'x measured downward from equilibrium'. Released 'above equilibrium' means x(0) is negative. The check in step 5 confirms x(0) = -1/2. This is consistent. The solution is sound.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-18 — The sign convention is inconsistent. The problem states the mass is released 1/2 ft *above* equilibrium, which corresponds to an initial position of x(0) = -1/2 if x is measured downward. However, the stated answer x(t) = -cos(4√6 t)/2 yields x(0) = -1/2, which is correct for the downward convention. But wait, let's re-read carefully. If x is measured downward, 'above equilibrium' is negative. So x(0) = -1/2. The solution checks out algebraically. Let me double check the physics. k = 192, m = 2. omega = sqrt(192/2) = sqrt(96) = 4*sqrt(6). Correct. x(t) = A cos(omega t) + B sin(omega t). x(0) = -1/2 => A = -1/2. x'(0) = 0 => B = 0. So x(t) = -1/2 cos(4*sqrt(6) t). The solution is actually correct. Why did I think it was wrong? Ah, I need to check if the sentence in step 3 defines the direction correctly. 'x measured downward from equilibrium'. Released 'above equilibrium' means x(0) is negative. The check in step 5 confirms x(0) = -1/2. This is consistent. The solution is sound.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the parameters, sets up the differential equation with the correct sign convention (downward positive), and verifies that the proposed solution satisfies both the equation and the initial conditions.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.