Simple harmonic motion worksheet answers for AP Physics 1 (free PDF download)

Problem 1 Which motions are SHM?

1. A and C are SHM (C only approximately, because the angle is small). In B the force changes abruptly at the floor and is constant in flight. In D the motion does not move back and forth along a line about an equilibrium point.
2. The net force must be proportional to the displacement from equilibrium and directed back toward equilibrium, F = −kx.

Problem 2 Metronome timing

1. f = 72 ÷ 60 s = 1.2 Hz.
2. T = 1 ÷ f = 1 ÷ 1.2 = 0.83 s.
3. N = 72 × 5.0 = 360 cycles.

Problem 3 Spring force

1. F = kx = 120 × 0.15 = 18 N.
2. The force is opposite to the displacement, pulling the end back toward the unstretched position.
3. F = 120 × 0.050 = 6.0 N, directed toward the equilibrium position.

Problem 4 Period of a spring

1. T = 2π√(m/k) = 2π√(0.60 ÷ 150) = 2π × 0.0632 = 0.40 s.
2. f = 1 ÷ 0.3974 = 2.5 Hz.
3. The period is unchanged at 0.40 s, because the period does not depend on amplitude.

Problem 5 Period of a pendulum

1. T = 2π√(L/g) = 2π√(0.640 ÷ 9.81) = 2π × 0.2554 = 1.60 s.
2. 1.60 s, because the mass cancels from the pendulum equation.

Problem 6 Reading an x–t graph

1. The amplitude is 6.0 cm.
2. One cycle takes 0.80 s, so f = 1 ÷ 0.80 = 1.25 Hz.
3. The speed is zero at the peaks: t = 0 and t = 0.40 s (and 0.80 s). The speed is greatest as the graph crosses x = 0: t = 0.20 s (or 0.60 s).

Problem 7 Spring constant from stretch

1. m = 0.350 kg and x = 0.084 m.
2. At equilibrium kx = mg, so k = 0.350 × 9.81 ÷ 0.084 = 40.9 N/m.
3. T = 2π√(0.350 ÷ 40.875) = 2π × 0.0925 = 0.58 s. (Check: T = 2π√(x/g) = 2π√(0.084 ÷ 9.81) = 0.58 s.)

Problem 8 Energy of a glider

1. E = ½kA² = ½ × 150 × 0.20² = 3.0 J.
2. At x = 0, ½mv² = E, so v = √(2 × 3.0 ÷ 0.80) = 2.7 m/s.
3. U = ½ × 150 × 0.10² = 0.75 J, so K = 3.0 − 0.75 = 2.25 J and v = √(2 × 2.25 ÷ 0.80) = 2.4 m/s.

Problem 9 Scaling the period

T = 2π√(m/k).
1. T is proportional to √m, so it doubles: 2.4 s.
2. The period does not depend on amplitude: 1.2 s.
3. T is proportional to 1/√k, so it halves: 0.60 s.
4. The mass and the spring constant are unchanged and gravity plays no part horizontally: 1.2 s.

Problem 10 Finding g with a pendulum

1. T = 33.0 ÷ 15 = 2.20 s.
2. T² = 4π²L/g, so g = 4π²L/T².
3. g = 39.48 × 1.20 ÷ 2.20² = 47.37 ÷ 4.84 = 9.79 m/s².
4. (9.81 − 9.79) ÷ 9.81 × 100 = 0.2 %.

Problem 11 Greatest acceleration

1. F = kA = 64 × 0.050 = 3.2 N.
2. a = F ÷ m = 3.2 ÷ 0.25 = 12.8 m/s².
3. Toward the equilibrium position, that is in the negative x direction, because a = −(k/m)x.

Problem 12 A vertical spring

1. mg = kx, so x = 0.40 × 9.81 ÷ 32 = 0.123 m.
2. T = 2π√(0.40 ÷ 32) = 2π × 0.1118 = 0.70 s.
3. The net force is kA = 32 × 0.050 = 1.6 N upward. The spring force is k(0.1226 + 0.050) = 5.5 N upward, which exceeds the weight of 3.9 N by 1.6 N.

Problem 13 Reading an energy graph

1. At x = ±0.10 m all energy is stored in the spring, so E = 1.0 J.
2. E = ½kA², so k = 2 × 1.0 ÷ 0.10² = 200 N/m.
3. U = ½ × 200 × 0.060² = 0.36 J, so K = 1.0 − 0.36 = 0.64 J.
4. v = √(2 × 0.64 ÷ 0.50) = 1.6 m/s.

Problem 14 Force and work from a graph

1. k = ΔF ÷ Δx = 18 ÷ 0.060 = 300 N/m.
2. The energy is the area under the line: ½ × 0.060 × 18 = 0.54 J.
3. The shaded trapezium has parallel sides 12 N and 18 N and width 0.020 m: ½ × (12 + 18) × 0.020 = 0.30 J. (Check: 0.54 − ½ × 300 × 0.040² = 0.54 − 0.24 = 0.30 J.)

Problem 15 Pendulum length for a target period

1. L = gT²/(4π²) = 9.81 × 4.00 ÷ 39.48 = 0.994 m.
2. L = 9.81 × 1.00 ÷ 39.48 = 0.248 m.
3. T is proportional to √L, so halving T needs one quarter of the length (0.994 ÷ 4 = 0.249 m), not one half.

Problem 16 Translating between graphs

1. k = 4π²m/T² = 39.48 × 0.20 ÷ 0.16 = 49 N/m.
2. v_max = 2πA/T = 2π × 0.050 ÷ 0.40 = 0.79 m/s. a_max = kA/m = 49.35 × 0.050 ÷ 0.20 = 12 m/s².
3. The x–t graph is a cosine, so the v–t graph is an inverted sine: it starts at 0, falls to −0.79 m/s at 0.10 s, returns to 0 at 0.20 s, rises to +0.79 m/s at 0.30 s and is 0 at 0.40 s. The a–t graph is an inverted cosine: it starts at −12 m/s², crosses zero at 0.10 s, reaches +12 m/s² at 0.20 s, crosses zero at 0.30 s and returns to −12 m/s² at 0.40 s.

Problem 17 Designing a timing experiment

1. 0.2 ÷ 1.5 × 100 = 13 %.
2. 20 swings take about 30 s, so 0.2 ÷ 30 × 100 = 0.7 %.
3. Independent variable: starting angle (for example 5°, 10°, 15°, 20°). Dependent variable: period T = time for 20 swings ÷ 20. Controlled: length, bob mass, same pivot and thread. Repeat each angle three times, average, and compare the periods; they should agree within the uncertainty for small angles.

Problem 18 Argue about a vertical spring

1. T = 2π√(0.50 ÷ 50) = 2π × 0.10 = 0.63 s.
2. x = mg/k = 0.50 × 9.81 ÷ 50 = 0.098 m.
3. Student 2 is correct. Gravity is a constant force, so it only moves the equilibrium position down by 0.098 m, where weight and spring force balance. Measured from there, the net force is −kx, the same restoring law as the horizontal case, so the acceleration pattern and the period 2π√(m/k) = 0.63 s are the same.

Problem 19 Clay dropped on a block

1. v = A√(k/m) = 0.30 × √(100 ÷ 0.90) = 3.16 m/s.
2. Horizontal momentum is conserved: v′ = 3.162 ÷ 2 = 1.58 m/s. Then ω′ = √(100 ÷ 1.80) = 7.454 rad/s and A′ = v′/ω′ = 1.581 ÷ 7.454 = 0.21 m (that is 0.30 ÷ √2).
3. T′ = 2π√(1.80 ÷ 100) = 0.84 s. Energy before: ½ × 100 × 0.30² = 4.5 J; after: ½ × 100 × 0.2121² = 2.25 J, so half the energy is lost in the inelastic collision.

Problem 20 Mass and period data

1. T² = 0.0986, 0.181, 0.263, 0.346, 0.428 s².
2. Gradient = (0.4277 − 0.0986) ÷ (0.50 − 0.10) = 0.823 s²/kg.
3. T² = (4π²/k)m, so k = 39.48 ÷ 0.823 = 48 N/m.
4. The line does not pass through the origin (intercept about 0.016 s²) because part of the spring’s own mass also oscillates, so the effective oscillating mass is slightly larger than the hanging mass.