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 is repeated oscillation where the acceleration is proportional to the displacement from equilibrium and points towards it, a = −ω²x. These worked answers correct the 20 problems of the Physics 11-12 worksheet on simple harmonic motion, from recognising SHM to putty on the block: 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. Compare your method as well as your final number, and check that your calculator is in radians for any sine or cosine.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB C.1
Problem 1 Recognising SHM
1. Only (A): acceleration is proportional to displacement with a minus sign, so it points towards the centre. (B) points away, (C) is not proportional to x, (D) is constant.2. ω² = 16, so ω = 4.0 rad/s. T = 2π/ω = 2π ÷ 4.0 = 1.57 s.
3. Towards the equilibrium position, that is in the negative direction.
Problem 2 Swing in the park
1. T = 36 ÷ 15 = 2.4 s.2. f = 1 ÷ T = 1 ÷ 2.4 = 0.417 Hz.
3. ω = 2πf = 2π × 0.4167 = 2.62 rad/s.
Problem 3 Phone vibration motor
1. ω = 2πf = 2π × 160 = 1005 rad/s (about 1.01 × 10³ rad/s).2. T = 1 ÷ 160 = 0.00625 s = 6.25 ms.
3. 160 × 0.25 = 40 oscillations.
Problem 4 Pendulum in Accra
1. T = 2π√(L/g) = 2π√(0.80 ÷ 9.81) = 2π × 0.2856 = 1.79 s.2. t = 25 × 1.794 = 44.9 s.
Problem 5 Spring balance
1. T = 2π√(m/k) = 2π√(0.40 ÷ 25) = 2π × 0.1265 = 0.795 s.2. It stays the same, because the period of SHM is independent of amplitude.
3. f = 1 ÷ 0.7948 = 1.26 Hz.
Problem 6 Reading a displacement graph
1. Amplitude = 8.0 cm.2. Period = 1.6 s, f = 1 ÷ 1.6 = 0.625 Hz.
3. Speed is greatest when the float passes x = 0 (steepest gradient), at 0.4 s and 1.2 s.
4. ω = 2π ÷ 1.6 = 3.927 rad/s. vmax = ωA = 3.927 × 0.080 = 0.314 m/s.
Problem 7 Maximum speed and acceleration
1. vmax = ωA = 20 × 0.030 = 0.60 m/s.2. amax = ω²A = 400 × 0.030 = 12 m/s².
3. Maximum speed at the equilibrium position; maximum acceleration at the ends of the motion, where the speed is momentarily zero.
Problem 8 Stiffness from the period
1. T = 2π√(m/k), so k = 4π²m/T².2. k = 4π² × 0.50 ÷ 0.90² = 19.74 ÷ 0.81 = 24.4 N/m.
3. T = 2π√(2.0 ÷ 24.37) = 2π × 0.2865 = 1.8 s (twice the original, since the mass is four times larger).
Problem 9 Measuring g with a pendulum
1. T = 40.0 ÷ 20 = 2.00 s.2. g = 4π²L/T² = 4π² × 0.995 ÷ 2.00² = 39.28 ÷ 4.00 = 9.82 m/s².
3. A reaction-time error of about 0.2 s is a much smaller fraction of 40 s than of 2 s, so the percentage uncertainty in T is reduced.
Problem 10 Speed at a given displacement
1. ω = 2π ÷ 0.80 = 7.854 rad/s.2. v = ω√(A² − x²) = 7.854 × √(5.0² − 3.0²) cm = 7.854 × 4.0 cm = 31.4 cm/s = 0.314 m/s.
3. vmax = ωA = 7.854 × 5.0 = 39.3 cm/s = 0.393 m/s, which is larger, as expected because 3.0 cm is not the centre.
Problem 11 Energy of a spring oscillator
1. E = ½kA² = ½ × 60 × 0.10² = 0.3 J.2. At the centre all energy is kinetic: ½mv² = E, so v = √(2E/m) = √(2 × 0.3 ÷ 0.30) = 1.41 m/s.
3. Ep = ½ × 60 × 0.060² = 0.108 J, so Ek = 0.3 − 0.108 = 0.192 J.
Problem 12 Reading an energy graph
1. Total energy = 0.40 J; amplitude = 0.20 m.2. E = ½kA², so k = 2E/A² = 2 × 0.40 ÷ 0.20² = 0.80 ÷ 0.040 = 20 N/m.
3. Ep = ½ × 20 × 0.10² = 0.10 J, so Ek = 0.40 − 0.10 = 0.30 J.
Problem 13 Hanging mass
1. At equilibrium kx = mg, so k = mg/e = 0.20 × 9.81 ÷ 0.049 = 40 N/m.2. T = 2π√(m/k) = 2π√(0.20 ÷ 40.04) = 0.444 s.
3. Since k = mg/e, m/k = e/g, so T = 2π√(e/g) = 2π√(0.049 ÷ 9.81) = 0.444 s, the same value.
Problem 14 Displacement at a given time
1. ω = 2π × 2.5 = 15.71 rad/s (5π).2. ωt = 15.71 × 0.050 = 0.7854 rad (π/4), so x = 4.0 cos(0.7854) = 2.83 mm.
3. x = 0 when ωt = π/2, so t = (π/2) ÷ 15.71 = 0.10 s (a quarter of the 0.40 s period).
Problem 15 Clock on the Moon
1. T ∝ 1/√g, so TMoon = 1.00 × √(9.81 ÷ 1.62) = 2.46 s.2. The mass-spring period is unchanged: T = 2π√(m/k) contains no g.
3. A pendulum is driven by the component of weight, which falls on the Moon. A spring force depends on extension only, so the restoring force at a given displacement stays the same.
Problem 16 Two springs combined
1. Parallel: each spring stretches by the same amount, so the forces add and k = 30 + 30 = 60 N/m.2. Series: each spring carries the full force and the extensions add, so 1/k = 1/30 + 1/30 and k = 15 N/m.
3. Parallel: T = 2π√(0.60 ÷ 60) = 2π × 0.100 = 0.628 s. Series: T = 2π√(0.60 ÷ 15) = 2π × 0.200 = 1.26 s.
Problem 17 Phone on a shaking shelf
1. The shelf can only push the phone upwards. With no push, the phone accelerates downwards at g (its weight alone). If the shelf accelerates downwards faster than g, the phone cannot follow it and loses contact.2. ω = 2π × 2.0 = 12.57 rad/s. Leaves when ω²A = g, so A = g/ω² = 9.81 ÷ 157.9 = 0.0155 m = 1.6 cm.
3. At the top of the motion, where the downward acceleration is greatest.
Problem 18 Driven oscillator data
1. The peaks are at f₀ = 2.0 Hz, so T = 1 ÷ 2.0 = 0.50 s.2. ω₀ = 2π × 2.0 = 12.57 rad/s. k = mω₀² = 0.25 × 157.9 = 39.5 N/m.
3. Heavier damping lowers the peak and makes it broader, but it stays near the same natural frequency.
4. Far above f₀ the driving force is out of step with the motion, so the amplitude is much smaller than at resonance (a small fraction of the peak value).
Problem 19 Decaying swings
1. A = 8.0 × 0.95¹⁰ = 8.0 × 0.5987 = 4.79 cm.2. E/E₀ = (0.95¹⁰)² = 0.3585, so energy lost = 1 − 0.3585 = 64.2 %.
3. 0.95²ⁿ = 0.5, so n = ln 0.5 ÷ (2 ln 0.95) = (−0.6931) ÷ (−0.10259) = 6.76 cycles.
Problem 20 Putty on the block
1. ω = √(200 ÷ 2.0) = 10 rad/s. v = ωA = 10 × 0.15 = 1.5 m/s.2. The putty lands vertically, so horizontal momentum is conserved: 2.0 × 1.5 = 3.0 × v, so v = 1 m/s.
3. New ω = √(200 ÷ 3.0) = 8.165 rad/s. Passing the centre, v = ωA, so A = 1 ÷ 8.165 = 0.122 m. New period T = 2π ÷ 8.165 = 0.77 s.
