Home›Calculus 3›Spring-mass systems›Problem 6.664

Spring-mass systems

Problem 6.664 · hard

A weight of \( \displaystyle 16 \) lb stretches a spring \( \displaystyle 2 \) ft. A damping force of \( \displaystyle 2 \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{2} \) ft below equilibrium with velocity \( \displaystyle -2 \) ft/s (down is positive). Find the position x(t).

\[ \frac{16}{2} = 8 \]
Hooke's law: k = weight / stretch (lb/ft).✓ Proved
\[ \frac{16}{32} = \frac{1}{2} \]
m = weight / g, with g = 32 ft/s².✓ Proved
The motion satisfies m x″ + c x′ + k x = 0: 1/2x″ + 2x′ + 8x = 0 (x measured downward from equilibrium).
Reviewed
\[ 2 \frac{d}{d t} \frac{\sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3} + \frac{\frac{d^{2}}{d t^{2}} \frac{\sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3}}{2} + \frac{8 \sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3} = 0 \]
This x(t) satisfies it ...✓ Proved
\[ \left. \frac{\sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3} \right|_{\substack{ t=0 }} = \frac{1}{2} \]
... starts in the right place ...✓ Proved
\[ \left. \frac{d}{d t} \frac{\sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3} \right|_{\substack{ t=0 }} = -2 \]
... with the right velocity.✓ Proved

Answer \( x(t) = \frac{\sqrt{3} e^{- 2 t} \cos{\left(2 \sqrt{3} t + \frac{\pi}{6} \right)}}{3} \)

✓ Nihil obstat Lines: 5 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. 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 differential equation parameters and verifies that the stated answer satisfies both the equation and the initial conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly derives the differential equation parameters and verifies that the stated answer satisfies both the equation and the initial conditions.
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly derives the differential equation parameters and verifies that the stated answer satisfies both the ODE and the initial conditions.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly derives the differential equation parameters and verifies that the stated answer satisfies both the ODE 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 2026-10-19 with SymPy 1.14.0.