Spring-mass systems
Problem 6.721 · hard
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 144 \) N/m. A damping force of \( \displaystyle 60 \) times the velocity acts on it. It is released from rest \( \displaystyle \frac{1}{2} \) m above equilibrium. Find the position x(t).
- The motion satisfies m x″ + c x′ + k x = 0: 4x″ + 60x′ + 144x = 0 (x measured downward from equilibrium).
- \[ 24 \left(1 - 4 e^{9 t}\right) e^{- 12 t} + 60 \frac{d}{d t} \frac{\left(1 - 4 e^{9 t}\right) e^{- 12 t}}{6} + 4 \frac{d^{2}}{d t^{2}} \frac{\left(1 - 4 e^{9 t}\right) e^{- 12 t}}{6} = 0 \]This x(t) satisfies it ...✓ Proved
- \[ \left. \frac{\left(1 - 4 e^{9 t}\right) e^{- 12 t}}{6} \right|_{\substack{ t=0 }} = - \frac{1}{2} \]... starts in the right place ...✓ Proved
- \[ \left. \frac{d}{d t} \frac{\left(1 - 4 e^{9 t}\right) e^{- 12 t}}{6} \right|_{\substack{ t=0 }} = 0 \]... with the right velocity.✓ Proved
Answer \( x(t) = \frac{\left(1 - 4 e^{9 t}\right) e^{- 12 t}}{6} \)
Lines: 3 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 | Not checked | — | 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: fail (style) — [domain objection, downgraded to style] The initial condition is stated as 1/2 m above equilibrium, which corresponds to x(0) = -1/2 given the downward-positive convention. However, the solution yields x(0) = -1/2, implying the mass starts 1/2 m below equilibrium (or the sign convention is inconsistent with the problem statement). Specifically, if x is measured downward, 'above equilibrium' means negative displacement. The solution gives x(0) = -1/2, which is correct for 'above'. Wait, let's re-read carefully. Problem: 'released from rest 1/2 m above equilibrium'. Convention: 'x measured downward'. Above equilibrium means x < 0. So x(0) should be -1/2. The solution checks out x(0) = -1/2. So the IC is correct. Let's check the ODE solution. Characteristic equation: 4r^2 + 60r + 144 = 0 => r^2 + 15r + 36 = 0 => (r+3)(r+12)=0. Roots are -3, -12. General solution: x(t) = c1 e^{-3t} + c2 e^{-12t}. x(0) = c1 + c2 = -1/2. x'(t) = -3c1 e^{-3t} - 12c2 e^{-12t}. x'(0) = -3c1 - 12c2 = 0 => c1 = -4c2. Sub into first: -4c2 + c2 = -1/2 => -3c2 = -1/2 => c2 = 1/6. c1 = -4/6 = -2/3. So x(t) = -2/3 e^{-3t} + 1/6 e^{-12t}. The stated answer is (1 - 4e^{9t})e^{-12t}/6 = 1/6 e^{-12t} - 4/6 e^{-3t} = 1/6 e^{-12t} - 2/3 e^{-3t}. This matches. The solution is actually correct. Why did I think it was wrong? Let me re-read the prompt's 'Stated answer'. x(t) = (1 - 4 e^{9 t}) e^{- 12 t}/6. This simplifies to 1/6 e^{-12t} - 2/3 e^{-3t}. My derivation matches. The setup is correct. The IC check is correct. The ODE check is correct. Is there a subtle error? 'A damping force of 60 times the velocity'. Usually damping is -cv. The equation is mx'' + cx' + kx = 0. This is standard. The direction of x is downward. Gravity is balanced by spring at equilibrium. So the homogeneous equation is correct. The initial position is 1/2 m above equilibrium. Downward is positive. So x(0) = -0.5. The solution has x(0) = -0.5. Everything seems correct. Wait, look at line 3. `Subs(..., t, 0) = -1/2`. The comment says `... starts in the right place ...`. The problem says '1/2 m above equilibrium'. If x is downward, above is negative. So -1/2 is correct. Is it possible the problem implies x is upward? 'x measured downward from equilibrium' is explicitly stated in line 1. So the sign is correct. Let me double check the roots. 4r^2 + 60r + 144 = 0. Divide by 4: r^2 + 15r + 36 = 0. (r+3)(r+12) = r^2 + 15r + 36. Correct. Roots -3, -12. Correct. Coefficients: c1 + c2 = -1/2. -3c1 - 12c2 = 0 => c1 = -4c2. -3c2 = -1/2 => c2 = 1/6. c1 = -4/6 = -2/3. x(t) = -2/3 e^{-3t} + 1/6 e^{-12t}. Stated answer: (1 - 4e^{9t})e^{-12t}/6 = 1/6 e^{-12t} - 4/6 e^{9t-12t} = 1/6 e^{-12t} - 2/3 e^{-3t}. Correct. The solution is sound. Why would it fail? Maybe the 'Stated answer' in the prompt is just a string, and the solution verifies it. The verification steps are marked [equation, proved]. The sentences are just setup and comments. The setup is correct. The comments are correct. I will mark it pass.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (style) 2026-10-20 — [domain objection, downgraded to style] The initial condition is stated as 1/2 m above equilibrium, which corresponds to x(0) = -1/2 given the downward-positive convention. However, the solution yields x(0) = -1/2, implying the mass starts 1/2 m below equilibrium (or the sign convention is inconsistent with the problem statement). Specifically, if x is measured downward, 'above equilibrium' means negative displacement. The solution gives x(0) = -1/2, which is correct for 'above'. Wait, let's re-read carefully. Problem: 'released from rest 1/2 m above equilibrium'. Convention: 'x measured downward'. Above equilibrium means x < 0. So x(0) should be -1/2. The solution checks out x(0) = -1/2. So the IC is correct. Let's check the ODE solution. Characteristic equation: 4r^2 + 60r + 144 = 0 => r^2 + 15r + 36 = 0 => (r+3)(r+12)=0. Roots are -3, -12. General solution: x(t) = c1 e^{-3t} + c2 e^{-12t}. x(0) = c1 + c2 = -1/2. x'(t) = -3c1 e^{-3t} - 12c2 e^{-12t}. x'(0) = -3c1 - 12c2 = 0 => c1 = -4c2. Sub into first: -4c2 + c2 = -1/2 => -3c2 = -1/2 => c2 = 1/6. c1 = -4/6 = -2/3. So x(t) = -2/3 e^{-3t} + 1/6 e^{-12t}. The stated answer is (1 - 4e^{9t})e^{-12t}/6 = 1/6 e^{-12t} - 4/6 e^{-3t} = 1/6 e^{-12t} - 2/3 e^{-3t}. This matches. The solution is actually correct. Why did I think it was wrong? Let me re-read the prompt's 'Stated answer'. x(t) = (1 - 4 e^{9 t}) e^{- 12 t}/6. This simplifies to 1/6 e^{-12t} - 2/3 e^{-3t}. My derivation matches. The setup is correct. The IC check is correct. The ODE check is correct. Is there a subtle error? 'A damping force of 60 times the velocity'. Usually damping is -cv. The equation is mx'' + cx' + kx = 0. This is standard. The direction of x is downward. Gravity is balanced by spring at equilibrium. So the homogeneous equation is correct. The initial position is 1/2 m above equilibrium. Downward is positive. So x(0) = -0.5. The solution has x(0) = -0.5. Everything seems correct. Wait, look at line 3. `Subs(..., t, 0) = -1/2`. The comment says `... starts in the right place ...`. The problem says '1/2 m above equilibrium'. If x is downward, above is negative. So -1/2 is correct. Is it possible the problem implies x is upward? 'x measured downward from equilibrium' is explicitly stated in line 1. So the sign is correct. Let me double check the roots. 4r^2 + 60r + 144 = 0. Divide by 4: r^2 + 15r + 36 = 0. (r+3)(r+12) = r^2 + 15r + 36. Correct. Roots -3, -12. Correct. Coefficients: c1 + c2 = -1/2. -3c1 - 12c2 = 0 => c1 = -4c2. -3c2 = -1/2 => c2 = 1/6. c1 = -4/6 = -2/3. x(t) = -2/3 e^{-3t} + 1/6 e^{-12t}. Stated answer: (1 - 4e^{9t})e^{-12t}/6 = 1/6 e^{-12t} - 4/6 e^{9t-12t} = 1/6 e^{-12t} - 2/3 e^{-3t}. Correct. The solution is sound. Why would it fail? Maybe the 'Stated answer' in the prompt is just a string, and the solution verifies it. The verification steps are marked [equation, proved]. The sentences are just setup and comments. The setup is correct. The comments are correct. I will mark it pass.gpt-oss:20b: pass 2026-10-20qwen3.6:27b-mlx: fail (error) 2026-10-19 — The initial condition is stated as 1/2 m above equilibrium, which corresponds to x(0) = -1/2 given the downward-positive convention. However, the stated answer yields x(0) = -1/2, but the problem implies the mass is released from rest *above* equilibrium. Wait, let's re-read carefully. 'Released from rest 1/2 m above equilibrium'. If x is measured downward, above equilibrium is negative. So x(0) = -1/2. The solution checks x(0) = -1/2. This seems consistent. Let's check the differential equation solution. Characteristic equation: 4r^2 + 60r + 144 = 0 => r^2 + 15r + 36 = 0 => (r+3)(r+12) = 0. Roots are -3 and -12. General solution: x(t) = c1 e^{-3t} + c2 e^{-12t}. x(0) = c1 + c2 = -1/2. x'(t) = -3c1 e^{-3t} - 12c2 e^{-12t}. x'(0) = -3c1 - 12c2 = 0 => c1 = -4c2. Substituting into x(0): -4c2 + c2 = -1/2 => -3c2 = -1/2 => c2 = 1/6. Then c1 = -4/6 = -2/3. So x(t) = -2/3 e^{-3t} + 1/6 e^{-12t}. The stated answer is (1 - 4e^{9t})e^{-12t}/6 = 1/6 e^{-12t} - 4/6 e^{-3t} = 1/6 e^{-12t} - 2/3 e^{-3t}. This matches. Why did I think it failed? Let's re-read the prompt's stated answer: x(t) = (1 - 4 e^{9 t}) e^{- 12 t} / 6. This simplifies to 1/6 e^{-12t} - 2/3 e^{-3t}. This is the correct solution. The solution steps are: 1. Setup ODE. 2. Check ODE. 3. Check x(0). 4. Check x'(0). All steps are correct and the answer is correct. Wait, is there a subtle error? 'x measured downward from equilibrium'. '1/2 m above equilibrium' -> x(0) = -0.5. The check says Subs(..., 0) = -1/2. This is correct. The derivative check is 0. This is correct. The ODE check is correct. The verdict should be pass.gpt-oss:20b: fail (error) 2026-10-19 — The proposed solution does not satisfy the initial displacement; the coefficient of the e^{-6t} term is incorrect. The correct solution should be x(t)= -e^{-3t}+0.5e^{-6t}.qwen3.6:27b-mlx: inconclusive 2026-10-19 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The initial condition is stated as 1/2 m above equilibrium, which corresponds to x(0) = -1/2 given the downward-positive convention. However, the progpt-oss:20b: fail (error) 2026-10-19 — The proposed solution does not satisfy the differential equation with the given parameters. The correct characteristic roots are r=−3 and r=−6, leading to x(t)=−e^(−3t)+½e^(−6t), not the expression provided. The initial conditions are also not met by the given formula.
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 2026-10-20 with SymPy 1.14.0.