The problems are on the simple harmonic motion worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Simple harmonic motion obeys d²x/dt² = −ω²x, with solution x = A cos(ωt + φ). These worked answers correct the 20 problems of the AP Physics C worksheet on simple harmonic motion, from reading an equation to motion from a potential energy: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit in bold. When a derivation is required, check that your steps start from Newton’s second law and end with d²q/dt² = −(constant)q. Round only at the end, and keep one extra digit in intermediate steps.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 7
Problem 1 Reading an equation
1. Comparing with x = A cos ωt: A = 0.12 m and ω = 8.0 rad/s.2. T = 2π/ω = 6.283 / 8.0 = 0.79 s.
3. f = 1/T = 1.3 Hz.
Problem 2 Spring and block
1. ω = √(k/m) = √(80 / 0.20) = √400 = 20 rad/s.2. T = 2π/ω = 6.283 / 20 = 0.31 s.
Problem 3 Pendulum on a cable
1. T = 2π√(L/g) = 2π√(2.50 / 9.81) = 2π × 0.5048 = 3.2 s.2. The mass cancels in √(L/g), so the period is unchanged.
Problem 4 Energy of a spring system
1. E = ½kA² = ½ × 300 × 0.050² = 0.375 J.2. At x = 0 the potential energy is zero, so the kinetic energy is 0.375 J.
Problem 5 Sign of the acceleration
1. a = −ω²x is negative, pointing back towards x = 0.2. At x = 0 the acceleration is zero and the speed is at its maximum.
3. The minus sign means a and x always have opposite signs, so the acceleration always points towards equilibrium.
Problem 6 Scaling the period
T = 2π√(m/k).1. √4 = 2, so the period becomes 2T.
2. The period does not depend on amplitude, so it stays T.
3. The factor is 1/√4, so the period becomes T/2.
Problem 7 Differentiating x(t)
1. v = dx/dt = −1.0 sin(5.0t) m/s and a = dv/dt = −5.0 cos(5.0t) m/s².2. At t = 0.30 s the angle is 1.5 rad, cos 1.5 = 0.0707, sin 1.5 = 0.9975. So x = 0.014 m, v = −1.0 m/s and a = −0.35 m/s².
Problem 8 Finding A and φ
1. A = √(x₀² + (v₀/ω)²) = √(0.0009 + 0.0016) = 0.050 m.2. cosφ = x₀/A = 0.60 and sinφ = −v₀/(Aω) = −0.80, so φ is in the fourth quadrant: φ = −0.93 rad (−53°).
Problem 9 Speed at a given position
1. ω = √(160 / 0.40) = 20 rad/s.2. v = ω√(A² − x²) = 20 × √(0.0100 − 0.0036) = 20 × 0.080 = 1.6 m/s.
Problem 10 Vertical spring
1. At rest kΔ = mg, so k = 0.30 × 9.81 / 0.060 = 49 N/m.2. T = 2π√(m/k) = 2π√(0.30 / 49.05) = 0.49 s. Gravity only shifts the equilibrium; it does not change T.
Problem 11 Graph reading
1. A = 4.0 cm and T = 0.80 s.2. ω = 2π/0.80 = 7.85 rad/s, so vmax = Aω = 0.040 × 7.85 = 0.31 m/s.
3. amax = Aω² = 0.040 × 61.7 = 2.5 m/s².
Problem 12 Two springs
1. k = 60 + 90 = 150 N/m, T = 2π√(0.50 / 150) = 0.36 s.2. k = (60 × 90) / 150 = 36 N/m, T = 2π√(0.50 / 36) = 0.74 s.
Problem 13 Equal kinetic and potential energy
KE = PE means ½kx² = ½·½kA², so x² = A²/2 and x = A/√2 = 0.14 / 1.414 = ±0.099 m.Problem 14 Spot the error
1. The formula comes from sinθ ≈ θ, which holds only for small angles (below about 10°).2. At 60° the restoring torque is weaker than the linear approximation, so the period is about 7 % longer than at 5°.
Problem 15 Pendulum on the Moon
1. T = 2π√(0.80 / 9.81) = 1.79 s.2. T = 2π√(0.80 / 1.62) = 4.42 s, which is √(9.81/1.62) = 2.46 times longer.
Problem 16 Rod as a physical pendulum
1. I = ⅓ML², d = 0.50 m: T = 2π√(⅓ / (9.81 × 0.50)) = 1.64 s.2. I = (1/12)ML² + Md² = M(0.0833 + 0.0625) = 0.1458 M; T = 2π√(0.1458 / (9.81 × 0.25)) = 1.53 s.
3. Pivot B gives the shorter period: the ratio I/(Md) is 0.667 m at A but only 0.583 m at B, and T = 2π√((I/Md)/g).
Problem 17 Disc pivoted on its rim
1. Parallel-axis theorem: I = ½MR² + MR² = 3MR²/2.2. ω = √(Mg R / (3MR²/2)) = √(2g / 3R) = √(2 × 9.81 / 0.60) = 5.7 rad/s, so T = 2π/ω = 1.1 s.
Problem 18 Torsion pendulum
1. T = 2π√(I/κ), so κ = 4π²I / T² = 39.48 × 0.0040 / 1.44 = 0.11 N m/rad.2. ω = 2π/1.2 = 5.24 rad/s, so θ̇max = 0.25 × 5.24 = 1.3 rad/s.
3. E = ½κθ₀² = ½ × 0.1097 × 0.0625 = 3.4 × 10⁻³ J.
Problem 19 Checking a solution
1. dx/dt = 0.60 cos 4.0t and d²x/dt² = −2.4 sin 4.0t = −(16)(0.15 sin 4.0t) = −16x, so ω² = 16 and ω = 4.0 rad/s.2. k = mω² = 2.0 × 16 = 32 N/m.
3. Fmax = kA = 32 × 0.15 = 4.8 N and E = ½kA² = ½ × 32 × 0.0225 = 0.36 J.
Problem 20 Motion from a potential energy
1. F = −dU/dx = −4.0x N, so k = 4.0 N/m.2. F = −kx, so the motion is SHM with ω = √(4.0 / 0.50) = 2.83 rad/s and T = 2π/ω = 2.2 s.
3. At the turning point E = U = 2.0A², so A = √(0.080 / 2.0) = 0.20 m.
