Spring-mass systems
Problem 6.497 · hard
A \( \displaystyle 4 \) kg mass hangs from a spring with constant \( \displaystyle 100 \) N/m. A damping force of \( \displaystyle 50 \) 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″ + 50x′ + 100x = 0 (x measured downward from equilibrium).
- \[ 50 \frac{d}{d t} \left(\frac{e^{- 10 t}}{6} - \frac{2 e^{- \frac{5 t}{2}}}{3}\right) + 4 \frac{d^{2}}{d t^{2}} \left(\frac{e^{- 10 t}}{6} - \frac{2 e^{- \frac{5 t}{2}}}{3}\right) + \frac{50 e^{- 10 t}}{3} - \frac{200 e^{- \frac{5 t}{2}}}{3} = 0 \]This x(t) satisfies it ...✓ Proved
- \[ \left. - \frac{2 e^{- \frac{5 t}{2}}}{3} + \frac{e^{- 10 t}}{6} \right|_{\substack{ t=0 }} = - \frac{1}{2} \]... starts in the right place ...✓ Proved
- \[ \left. \frac{d}{d t} \left(- \frac{2 e^{- \frac{5 t}{2}}}{3} + \frac{e^{- 10 t}}{6}\right) \right|_{\substack{ t=0 }} = 0 \]... with the right velocity.✓ Proved
Answer \( x(t) = \frac{e^{- 10 t}}{6} - \frac{2 e^{- \frac{5 t}{2}}}{3} \)
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 (error) — The initial condition is stated as '1/2 m above equilibrium' with x measured downward, which implies x(0) = -1/2. However, the proposed solution yields x(0) = 1/6 - 2/3 = -1/2, which matches. Wait, let's re-read the solution's claim. The solution claims the answer is x(t) = e^(-10t)/6 - 2e^(-5t/2)/3. Let's check the roots of the characteristic equation: 4r^2 + 50r + 100 = 0 => 2r^2 + 25r + 50 = 0. Roots are r = [-25 +/- sqrt(625 - 400)] / 4 = [-25 +/- 15] / 4. So r1 = -10/4 = -2.5 and r2 = -40/4 = -10. The general solution is x(t) = c1 e^(-2.5t) + c2 e^(-10t). The proposed solution has terms e^(-10t) and e^(-2.5t). Let's check coefficients. x(0) = c1 + c2 = -1/2. x'(0) = -2.5c1 - 10c2 = 0 => c1 = -4c2. So -4c2 + c2 = -1/2 => -3c2 = -1/2 => c2 = 1/6. Then c1 = -4/6 = -2/3. So x(t) = -2/3 e^(-2.5t) + 1/6 e^(-10t). This matches the stated answer. The solution checks out algebraically. Why did I think it failed? Let me re-read the prompt's 'Stated answer'. x(t) = e^(-10t)/6 - 2e^(-5t/2)/3. This is exactly what I derived. The solution steps are: 1. Setup ODE. 2. Verify ODE. 3. Verify x(0). 4. Verify x'(0). All steps are correct. The verdict should be pass.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The initial condition is stated as '1/2 m above equilibrium' with x measured downward, which implies x(0) = -1/2. However, the proposed solution yields x(0) = 1/6 - 2/3 = -1/2, which matches. Wait, let's re-read the solution's claim. The solution claims the answer is x(t) = e^(-10t)/6 - 2e^(-5t/2)/3. Let's check the roots of the characteristic equation: 4r^2 + 50r + 100 = 0 => 2r^2 + 25r + 50 = 0. Roots are r = [-25 +/- sqrt(625 - 400)] / 4 = [-25 +/- 15] / 4. So r1 = -10/4 = -2.5 and r2 = -40/4 = -10. The general solution is x(t) = c1 e^(-2.5t) + c2 e^(-10t). The proposed solution has terms e^(-10t) and e^(-2.5t). Let's check coefficients. x(0) = c1 + c2 = -1/2. x'(0) = -2.5c1 - 10c2 = 0 => c1 = -4c2. So -4c2 + c2 = -1/2 => -3c2 = -1/2 => c2 = 1/6. Then c1 = -4/6 = -2/3. So x(t) = -2/3 e^(-2.5t) + 1/6 e^(-10t). This matches the stated answer. The solution checks out algebraically. Why did I think it failed? Let me re-read the prompt's 'Stated answer'. x(t) = e^(-10t)/6 - 2e^(-5t/2)/3. This is exactly what I derived. The solution steps are: 1. Setup ODE. 2. Verify ODE. 3. Verify x(0). 4. Verify x'(0). All steps are correct. The verdict should be pass.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (error) 2026-10-15 — The initial condition is stated as 1/2 m above equilibrium, which corresponds to x(0) = -1/2 if x is measured downward. However, the provided solution yields x(0) = -1/2, but the problem likely implies the standard convention where 'above' might be negative or positive depending on axis definition. Let's re-read carefully: 'released from rest 1/2 m above equilibrium'. If x is downward, above is negative, so x(0) = -0.5. The solution checks out for x(0)=-0.5. Wait, let's look at the differential equation coefficients. m=4, c=50, k=100. Characteristic equation: 4r^2 + 50r + 100 = 0 => 2r^2 + 25r + 50 = 0. Roots: r = [-25 +/- sqrt(625 - 400)]/4 = [-25 +/- 15]/4. r1 = -10/4 = -2.5, r2 = -40/4 = -10. So the general solution is x(t) = c1 e^{-2.5t} + c2 e^{-10t}. The stated answer has terms e^{-5t/2} (which is e^{-2.5t}) and e^{-10t}. This matches the roots. Let's check the constants. x(0) = c1 + c2 = -1/2. x'(t) = -2.5 c1 e^{-2.5t} - 10 c2 e^{-10t}. x'(0) = -2.5 c1 - 10 c2 = 0 => c1 = -4 c2. Substitute into first eq: -4c2 + c2 = -1/2 => -3c2 = -1/2 => c2 = 1/6. Then c1 = -4/6 = -2/3. So x(t) = -2/3 e^{-2.5t} + 1/6 e^{-10t}. The stated answer is 1/6 e^{-10t} - 2/3 e^{-2.5t}. This matches. The solution steps are correct. Why did I think it was wrong? Let me re-read the prompt's 'Stated answer'. x(t) = e^{-10t}/6 - 2e^{-5t/2}/3. This is exactly what I derived. The setup is correct. The verification steps are correct. Is there a subtle error? 'x measured downward'. '1/2 m above equilibrium'. So x(0) = -0.5. The solution uses x(0) = -1/2. This is consistent. The verdict should be pass.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution claims the initial position is -1/2 m, but the problem states the mass is released 1/2 m above equilibrium. With x measured downward, this corresponds to x(0) = -1/2, which matches the check, but the stated answer x(t) = e^(-10t)/6 - 2e^(-5t/2)/3 yields x(0) = 1/6 - 2/3 = -1/2. Wait, let's re-read carefully. The problem says 'released from rest 1/2 m above equilibrium'. If x is downward, above equilibrium is negative x. So x(0) = -1/2. The check says Subs(..., 0) = -1/2. This is consistent. However, let's look at the differential equation solution. Characteristic equation: 4r^2 + 50r + 100 = 0 => 2r^2 + 25r + 50 = 0. Roots: r = [-25 +/- sqrt(625 - 400)] / 4 = [-25 +/- 15] / 4. r1 = -10/4 = -2.5, r2 = -40/4 = -10. So x(t) = c1 e^(-2.5t) + c2 e^(-10t). x(0) = c1 + c2 = -1/2. x'(0) = -2.5c1 - 10c2 = 0 => c1 = -4c2. -4c2 + c2 = -1/2 => -3c2 = -1/2 => c2 = 1/6. c1 = -4/6 = -2/3. So x(t) = -2/3 e^(-2.5t) + 1/6 e^(-10t). The stated answer is e^(-10t)/6 - 2e^(-5t/2)/3. This matches. Why is it a fail? Let's check the sentence 1. 'x measured downward from equilibrium'. If x is downward, gravity is balanced by spring force at equilibrium. The equation mx'' + cx' + kx = 0 is correct for displacement from equilibrium. The initial condition is '1/2 m above equilibrium'. Above equilibrium means negative x if x is downward. So x(0) = -1/2. The solution checks out. Is there a subtle error? 'A damping force of 50 times the velocity acts on it.' Usually damping opposes motion, so F_d = -cv. The term in the ODE is +cx' on the LHS, so 50x'. This is correct. The roots are correct. The constants are correct. The answer is correct. Why would I fail it? Maybe the sign convention? If x is upward, x(0) = 1/2. Then x(t) = 2/3 e^(-2.5t) - 1/6 e^(-10t). The given answer is the negative of this. The sentence says 'x measured downward'. This is a valid choice. The answer is consistent with this choice. It seems correct. Let me double check the 'unchecked' equations. They are just verification steps. The prompt asks to review the solution. The solution provides the answer and verifies it. Is the verification sufficient? It checks the ODE and ICs. That is sufficient for uniqueness. Is there a style issue? 'Subs(...)' is not standard math notation, but it's marked as an equation checked by a CAS. The sentence 1 is the main content. It sets up the model correctly. I will pass it.gpt-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.