A 0.50 kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. At what location are the kinetic energy and the potential energy the same?

Question

BlackTom AI · Accepted Answer

Total mechanical energy in simple harmonic motion is conserved and equals kinetic energy at the equilibrium point. At x = 0 the spring potential is zero, so E = K_max = 1/2 m v^2. We want the position x where K = U, so each equals half the total energy.

Relevant concepts:
- Total energy E = K + U is constant.
- K = 1/2 m v^2, U = 1/2 k x^2.
- At equilibrium v = vmax and U = 0, so E = 1/2 m vmax^2.
- If K = U then K = U = E/2.

Step-by-step calculation:
1) Compute total energy using given mass and speed at equilibrium:
   E = 1/2 m v^2 = 0.5 × 0.50 kg × (1.5 m/s)^2
   (1.5)^2 = 2.25, so E = 0.5 × 0.50 × 2.25 = 0.5625 J.

2) When K = U, each equals E/2:
   U = E/2 = 0.5625 J / 2 = 0.28125 J.

3) Set spring potential energy equal to this value and solve for x:
   1/2 k x^2 = 0.28125 J
   x^2 = (2 × 0.28125) / k = 0.5625 / 20 = 0.028125
   x = sqrt(0.028125) ≈ 0.1677 m ≈ 0.17 m.

Alternate quick method: amplitude A satisfies E = 1/2 k A^2 = 1/2 m v^2, and the position where K = U is x = A/√2. Using ω = sqrt(k/m) gives A = vmax/ω = 1.5 / sqrt(40) ≈ 0.2372 m, so x = 0.2372/√2 ≈ 0.1677 m, same result.

Why the correct answer is correct:
At x ≈ 0.17 m half the total mechanical energy is stored as spring potential and half as kinetic, matching the condition K = U.

Why other options are incorrect:
- 0.017 m and 0.029 m are an order of magnitude too small (would give much smaller U than required).
- 0.12 m yields U = 1/2·20·(0.12)^2 = 0.144 J, which is less than the required 0.281 J.

Final answer confirmed: 0.17 m.

A 0.50 kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. At what location are the kinetic energy and the potential energy the same?单项选择题

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