Spring-mass systems
Problem 6.660 · hard
A \( \displaystyle 2 \) kg mass hangs from a spring with constant \( \displaystyle 50 \) N/m. A damping force of \( \displaystyle 25 \) times the velocity acts on it. It starts \( \displaystyle \frac{1}{2} \) m above equilibrium with velocity \( \displaystyle -2 \) m/s (down is positive). Find the position x(t).
- The motion satisfies m x″ + c x′ + k x = 0: 2x″ + 25x′ + 50x = 0 (x measured downward from equilibrium).
- \[ 25 \frac{d}{d t} \left(\frac{13 e^{- 10 t}}{30} - \frac{14 e^{- \frac{5 t}{2}}}{15}\right) + 2 \frac{d^{2}}{d t^{2}} \left(\frac{13 e^{- 10 t}}{30} - \frac{14 e^{- \frac{5 t}{2}}}{15}\right) + \frac{65 e^{- 10 t}}{3} - \frac{140 e^{- \frac{5 t}{2}}}{3} = 0 \]This x(t) satisfies it ...✓ Proved
- \[ \left. - \frac{14 e^{- \frac{5 t}{2}}}{15} + \frac{13 e^{- 10 t}}{30} \right|_{\substack{ t=0 }} = - \frac{1}{2} \]... starts in the right place ...✓ Proved
- \[ \left. \frac{d}{d t} \left(- \frac{14 e^{- \frac{5 t}{2}}}{15} + \frac{13 e^{- 10 t}}{30}\right) \right|_{\substack{ t=0 }} = -2 \]... with the right velocity.✓ Proved
Answer \( x(t) = \frac{13 e^{- 10 t}}{30} - \frac{14 e^{- \frac{5 t}{2}}}{15} \)
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: fail (error) — The characteristic equation for 2x″+25x′+50x=0 gives roots r=−5 and r=−2.5, so the general solution should be a linear combination of e^(−5t) and e^(−2.5t). The proposed x(t)=13/30 e^(−10t)−14/15 e^(−5t/2) uses exponents −10 and −2.5, which do not satisfy the differential equation, making the solution incorrect.qwen3.6:27b-mlx: fail (error) — The initial condition x(0) is stated as 1/2 m above equilibrium, which corresponds to x(0) = -1/2 given the downward-positive convention. However, the provided solution yields x(0) = -1/2, but the problem states the mass starts 1/2 m *above* equilibrium. Wait, let's re-read carefully. 'starts 1/2 m above equilibrium'. Down is positive. So x(0) should be -0.5. The solution checks out for x(0)=-0.5. Let's check the velocity. 'velocity -2 m/s'. The solution checks out for v(0)=-2. Let's check the differential equation. m=2, c=25, k=50. Characteristic equation: 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^{-10t} and e^{-2.5t} (since 5/2 = 2.5). Let's check the coefficients. x(0) = c1 + c2 = -1/2. x'(0) = -2.5 c1 - 10 c2 = -2. Multiply first by 2.5: 2.5 c1 + 2.5 c2 = -1.25. Add to second: -7.5 c2 = -3.25 => c2 = 3.25 / 7.5 = 13/30. Then c1 = -1/2 - 13/30 = -15/30 - 13/30 = -28/30 = -14/15. So x(t) = -14/15 e^{-2.5t} + 13/30 e^{-10t}. The stated answer is 13/30 e^{-10t} - 14/15 e^{-2.5t}. This matches. The setup and solution are correct. Why did I think it was wrong? Let me re-read the prompt's 'Stated answer'. x(t) = 13 e^{-10t}/30 - 14 e^{-5t/2}/15. This is exactly what I derived. The solution steps are: 1. Setup ODE. 2. Check ODE. 3. Check x(0). 4. Check x'(0). All steps are valid and correct. The verdict should be pass.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-19 — The initial condition x(0) is stated as 1/2 m above equilibrium, which corresponds to x(0) = -1/2 given the downward-positive convention. However, the provided solution yields x(0) = -1/2, but the problem states the mass starts 1/2 m *above* equilibrium. Wait, let's re-read carefully. 'starts 1/2 m above equilibrium'. Down is positive. So x(0) should be -0.5. The solution checks out for x(0)=-0.5. Let's check the velocity. 'velocity -2 m/s'. The solution checks out for v(0)=-2. Let's check the differential equation. m=2, c=25, k=50. Characteristic equation: 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^{-10t} and e^{-2.5t} (since 5/2 = 2.5). Let's check the coefficients. x(0) = c1 + c2 = -1/2. x'(0) = -2.5 c1 - 10 c2 = -2. Multiply first by 2.5: 2.5 c1 + 2.5 c2 = -1.25. Add to second: -7.5 c2 = -3.25 => c2 = 3.25 / 7.5 = 13/30. Then c1 = -1/2 - 13/30 = -15/30 - 13/30 = -28/30 = -14/15. So x(t) = -14/15 e^{-2.5t} + 13/30 e^{-10t}. The stated answer is 13/30 e^{-10t} - 14/15 e^{-2.5t}. This matches. The setup and solution are correct. Why did I think it was wrong? Let me re-read the prompt's 'Stated answer'. x(t) = 13 e^{-10t}/30 - 14 e^{-5t/2}/15. This is exactly what I derived. The solution steps are: 1. Setup ODE. 2. Check ODE. 3. Check x(0). 4. Check x'(0). All steps are valid and correct. The verdict should be pass.gpt-oss:20b: fail (error) 2026-10-19 — The characteristic equation for 2x″+25x′+50x=0 gives roots r=−5 and r=−2.5, so the general solution should be a linear combination of e^(−5t) and e^(−2.5t). The proposed x(t)=13/30 e^(−10t)−14/15 e^(−5t/2) uses exponents −10 and −2.5, which do not satisfy the differential equation, making the solution incorrect.qwen3.6:27b-mlx: fail (error) 2026-10-16 — The initial position is stated as 1/2 m above equilibrium, which corresponds to x(0) = -1/2 since down is positive. However, the provided solution yields x(0) = -1/2, but the stated answer in the problem description is x(t) = 13e^{-10t}/30 - 14e^{-5t/2}/15. Let's check the sign. If x(0) = -1/2, then 13/30 - 28/30 = -15/30 = -1/2. This matches. Wait, let me re-read the problem statement carefully. 'It starts 1/2 m above equilibrium'. Down is positive. So x(0) should be -1/2. The solution checks out for x(0)=-1/2. Is there an error? Let's check the velocity. x'(t) = -130/30 e^{-10t} + 70/30 e^{-5t/2}. x'(0) = -13/3 + 7/3 = -6/3 = -2. This matches v(0)=-2. The differential equation is 2x'' + 25x' + 50x = 0. Characteristic equation: 2r^2 + 25r + 50 = 0. Roots: r = [-25 +/- sqrt(625 - 400)] / 4 = [-25 +/- 15] / 4. r1 = -10/4 = -2.5 = -5/2. r2 = -40/4 = -10. So the general solution is c1 e^{-10t} + c2 e^{-2.5t}. The solution uses these exponents. The constants are c1 = 13/30 and c2 = -14/15 = -28/30. Sum = -15/30 = -1/2. Correct. Velocity: -10c1 - 2.5c2 = -130/30 - 2.5(-28/30) = -13/3 + 70/30 = -130/30 + 70/30 = -60/30 = -2. Correct. Why would this fail? Ah, look at the stated answer in the prompt: x(t) = 13e^{-10t}/30 - 14e^{-5t/2}/15. The solution verifies this. Is there a subtle error? 'A damping force of 25 times the velocity acts on it.' Usually damping force is -cv. The equation is mx'' = -kx - cv. So mx'' + cv + kx = 0. This is correct. 'Down is positive'. Gravity is balanced by spring extension at equilibrium. So x is displacement from equilibrium. This is standard. The setup is correct. The math is correct. The verification is correct. Is it possible the problem implies 'above equilibrium' means positive x? No, it explicitly says 'down is positive'. So above is negative. x(0) = -1/2. The solution gets x(0) = -1/2. Everything seems correct. Let me double check the 'Stated answer' vs the 'Solution'. The solution claims the stated answer is correct. It is. Why would I mark it fail? Maybe I missed a sign in the problem statement? 'starts 1/2 m above equilibrium'. If down is positive, x(0) = -0.5. The solution has x(0) = -0.5. 'velocity -2 m/s'. x'(0) = -2. The solution has x'(0) = -2. The DE is 2x'' + 25x' + 50x = 0. Roots -10, -2.5. Solution form A e^{-10t} + B e^{-2.5t}. A+B = -0.5. -10A - 2.5B = -2. Multiply first by 2.5: 2.5A + 2.5B = -1.25. Subtract from second: -12.5A = -0.75 => A = 0.75/12.5 = 3/50 = 6/100 = 0.06. Wait. 0.75 / 12.5 = 7.5 / 125 = 1.5 / 25 = 3/50. 3/50 = 9/150. The solution has 13/30 = 65/150. Something is wrong. Let's re-solve. A+B = -1/2. -10A - 5/2 B = -2. Multiply by 2: -20A - 5B = -4. From first: B = -1/2 - A. Sub into second: -20A - 5(-1/2 - A) = -4. -20A + 5/2 + 5A = -4. -15A = -4 - 2.5 = -6.5. A = 6.5 / 15 = 13/30. Correct. B = -1/2 - 13/30 = -15/30 - 13/30 = -28/30 = -14/15. Correct. The solution is actually correct. Why did I think it might be wrong? I miscalculated 3/50 earlier. 6.5/15 is indeed 13/30. Okay, so the solution is correct. Is there any ambiguity? 'Damping force of 25 times the velocity'. Force is a vector. If velocity is v, force is -25v (opposing motion). Equation: m a = F_spring + F_damping. F_spring = -kx. F_damping = -cv. m x'' = -kx - cv. m x'' + cv + kx = 0. Correct. The solution is sound. I will pass it.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-15 — The initial position is given as 1/2 m above equilibrium, which corresponds to x(0) = -1/2 since down is positive. However, the stated answer yields x(0) = 13/30 - 28/30 = -15/30 = -1/2, which is correct. Wait, let me re-read the check. Line 3 says Subs(..., t, 0) = -1/2. This matches the problem statement. Line 4 checks velocity. Let's check the velocity of the stated answer: x'(t) = -130/30 e^{-10t} + 70/30 e^{--5t/2} = -13/3 e^{-10t} + 7/3 e^{-5t/2}. At t=0, x'(0) = -13/3 + 7/3 = -6/3 = -2. This matches. So the stated answer is actually correct for the problem as stated. Why would it fail? Let me re-read the prompt's solution lines. Line 1 sets up the ODE. Line 2 checks the ODE. Line 3 checks x(0). Line 4 checks x'(0). All checks pass. Is there a subtle error? The problem says 'starts 1/2 m above equilibrium'. Down is positive. So x(0) = -0.5. The solution checks for -1/2. This is correct. The velocity is -2 m/s. The solution checks for -2. This is correct. The ODE is 2x'' + 25x' + 50x = 0. Characteristic equation: 2r^2 + 25r + 50 = 0. Roots: r = [-25 +/- sqrt(625 - 400)] / 4 = [-25 +/- 15] / 4. r1 = -10/4 = -2.5 = -5/2. r2 = -40/4 = -10. So the general solution is x(t) = c1 e^{-5t/2} + c2 e^{-10t}. The stated answer is x(t) = 13/30 e^{-10t} - 14/15 e^{-5t/2}. Let's check the coefficients. x(0) = 13/30 - 28/30 = -15/30 = -1/2. Correct. x'(0) = 13/30(-10) - 14/15(-5/2) = -130/30 + 70/30 = -60/30 = -2. Correct. The solution seems perfectly correct. Why would I mark it fail? Maybe I missed something in the 'adversarial' instruction. 'Find what is wrong'. If nothing is wrong, I should say pass. Let me double check the 'unchecked' lines. The prompt says 'Each line... is either an EQUATION... or a SENTENCE'. Lines 2, 3, 4 are marked [equation, unchecked]. The prompt says 'a computer algebra system has already checked (its result is shown; do not re-check the algebra of a line marked proved)'. Wait, the prompt says 'do not re-check the algebra of a line marked proved'. These lines are marked 'unchecked'. Does that mean I should check them? Or does it mean the CAS hasn't checked them? The prompt says 'Each line... is either an EQUATION, which a computer algebra system has already checked... or a SENTENCE'. This implies that if it's an EQUATION, it's checked. But the tags say 'unchecked'. This is a contradiction in the prompt format description vs the specific tags. However, usually 'unchecked' in these prompts means 'the CAS output is provided but you should verify the logic or the setup'. But here the lines are just verification steps. If the verification steps are correct, and the setup is correct, it's a pass. Let me look really closely at Line 1. '2x'' + 25x' + 50x = 0'. Mass 2, damping 25, spring 50. Correct. 'x measured downward'. Correct. 'starts 1/2 m above equilibrium'. x(0) = -1/2. Correct. 'velocity -2 m/s'. x'(0) = -2. Correct. The stated answer satisfies the ODE and ICs. It seems correct. Is there a convention issue? 'down is positive'. 'above equilibrium' is negative x. Correct. Velocity '-2 m/s'. If down is positive, negative velocity is upward. The problem states velocity is -2 m/s. It doesn't say 'upward', it gives the signed value. So x'(0) = -2. Correct. I see no error. I will mark it pass.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.