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Spring-mass systems

Problem 6.635 · medium

A weight of \( \displaystyle 64 \) lb stretches a spring \( \displaystyle 1 \) ft. There is no damping. It starts \( \displaystyle \frac{1}{2} \) ft above equilibrium with velocity \( \displaystyle 1 \) ft/s (down is positive). Find the position x(t).

\[ \frac{64}{1} = 64 \]
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′ + 64x = 0 (x measured downward from equilibrium).
Reviewed
\[ 8 \sqrt{2} \sin{\left(4 \sqrt{2} t \right)} - 32 \cos{\left(4 \sqrt{2} t \right)} + 2 \frac{d^{2}}{d t^{2}} \left(\frac{\sqrt{2} \sin{\left(4 \sqrt{2} t \right)}}{8} - \frac{\cos{\left(4 \sqrt{2} t \right)}}{2}\right) = 0 \]
This x(t) satisfies it ...✓ Proved
\[ \left. \frac{\sqrt{2} \sin{\left(4 \sqrt{2} t \right)}}{8} - \frac{\cos{\left(4 \sqrt{2} t \right)}}{2} \right|_{\substack{ t=0 }} = - \frac{1}{2} \]
... starts in the right place ...✓ Proved
\[ \left. \frac{d}{d t} \left(\frac{\sqrt{2} \sin{\left(4 \sqrt{2} t \right)}}{8} - \frac{\cos{\left(4 \sqrt{2} t \right)}}{2}\right) \right|_{\substack{ t=0 }} = 1 \]
... with the right velocity.✓ Proved

Answer \( x(t) = \frac{\sqrt{2} \sin{\left(4 \sqrt{2} t \right)}}{8} - \frac{\cos{\left(4 \sqrt{2} t \right)}}{2} \)

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: pass
qwen3.6:27b-mlx: pass — The solution correctly derives the parameters k and m, sets up the differential equation with the correct sign convention (down is positive), and verifies that the stated answer satisfies both the differential equation and the initial conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly derives the parameters k and m, sets up the differential equation with the correct sign convention (down is positive), and verifies that the stated answer satisfies both the differential equation and the initial conditions.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: inconclusive 2026-10-16 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The initial position is given as 1/2 ft above equilibrium, which corresponds to x(0) = -1/2 since down is positive. However, the stated answer yields
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The initial condition for position is incorrect. The problem states the weight starts 1/2 ft *above* equilibrium with *down* as positive, so x(0) sho
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.