Spring-mass systems
Problem 6.641 · hard
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 2 \) N/m. A damping force of \( \displaystyle \frac{12}{5} \) times the velocity acts on it. It is released from rest \( \displaystyle \frac{1}{3} \) m below equilibrium. Find the position x(t).
- The motion satisfies m x″ + c x′ + k x = 0: 2x″ + 12/5x′ + 2x = 0 (x measured downward from equilibrium).Reviewed
- \[ 2 \left(\frac{\sin{\left(\frac{4 t}{5} \right)}}{4} + \frac{\cos{\left(\frac{4 t}{5} \right)}}{3}\right) e^{- \frac{3 t}{5}} + \frac{12 \frac{d}{d t} \left(\frac{\sin{\left(\frac{4 t}{5} \right)}}{4} + \frac{\cos{\left(\frac{4 t}{5} \right)}}{3}\right) e^{- \frac{3 t}{5}}}{5} + 2 \frac{d^{2}}{d t^{2}} \left(\frac{\sin{\left(\frac{4 t}{5} \right)}}{4} + \frac{\cos{\left(\frac{4 t}{5} \right)}}{3}\right) e^{- \frac{3 t}{5}} = 0 \]This x(t) satisfies it ...✓ Proved
- \[ \left. \left(\frac{\sin{\left(\frac{4 t}{5} \right)}}{4} + \frac{\cos{\left(\frac{4 t}{5} \right)}}{3}\right) e^{- \frac{3 t}{5}} \right|_{\substack{ t=0 }} = \frac{1}{3} \]... starts in the right place ...✓ Proved
- \[ \left. \frac{d}{d t} \left(\frac{\sin{\left(\frac{4 t}{5} \right)}}{4} + \frac{\cos{\left(\frac{4 t}{5} \right)}}{3}\right) e^{- \frac{3 t}{5}} \right|_{\substack{ t=0 }} = 0 \]... with the right velocity.✓ Proved
Answer \( x(t) = \left(\frac{\sin{\left(\frac{4 t}{5} \right)}}{4} + \frac{\cos{\left(\frac{4 t}{5} \right)}}{3}\right) e^{- \frac{3 t}{5}} \)
✓ 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 — The solution correctly sets up the differential equation with the given parameters and verifies that the proposed solution satisfies both the equation and the initial conditions.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly sets up the differential equation with the given parameters and verifies that the proposed solution satisfies both the equation and the initial conditions.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The differential equation is correctly set up with the given parameters and initial conditions. The solution satisfies the ODE and the initial conditions as verified by the algebraic checks.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The setup correctly models the damped harmonic oscillator with the given parameters and initial conditions. The verification steps confirm the solution satisfies the differential equation and 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.