Spring-mass systems
Problem 6.491 · 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 starts \( \displaystyle \frac{1}{4} \) m below equilibrium with velocity \( \displaystyle 1 \) m/s (down is positive). Find the position x(t).
- The motion satisfies m x″ + c x′ + k x = 0: 2x″ + 10x′ + 50x = 0 (x measured downward from equilibrium).Reviewed
- \[ \frac{5 \left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{6} + 10 \frac{d}{d t} \frac{\left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{60} + 2 \frac{d^{2}}{d t^{2}} \frac{\left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{60} = 0 \]This x(t) satisfies it ...✓ Proved
- \[ \left. \frac{\left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{60} \right|_{\substack{ t=0 }} = \frac{1}{4} \]... starts in the right place ...✓ Proved
- \[ \left. \frac{d}{d t} \frac{\left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{60} \right|_{\substack{ t=0 }} = 1 \]... with the right velocity.✓ Proved
Answer \( x(t) = \frac{\left(13 \sqrt{3} \sin{\left(\frac{5 \sqrt{3} t}{2} \right)} + 15 \cos{\left(\frac{5 \sqrt{3} t}{2} \right)}\right) e^{- \frac{5 t}{2}}}{60} \)
✓ Nihil obstat Lines: 3 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly sets up the differential equation with the given parameters and initial conditions, and the provided answer is verified to satisfy both the ODE and the initial conditions.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: pass 2026-10-14gpt-oss:20b: pass 2026-10-14
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 2026-10-17 with SymPy 1.14.0.