A2 · Practice questions · Simple harmonic motion

Acceleration opposite to displacement.

Six original Cambridge-style questions on the defining condition a = −ω²x, the equations x = x₀ sin ωt and v = ±ω√(x₀² − x²), and how the three graphs line up.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

One condition, three equations.

01
Analysis
[2 marks]

A ball dropped onto a hard floor bounces repeatedly, rising to the same height each time. State the condition that must be met for a motion to be simple harmonic, and explain why the bouncing ball does not satisfy it.

  • For simple harmonic motion the acceleration must be proportional to the displacement from a fixed point and always directed towards that point, a = −ω²x. ✓
  • Between bounces the ball's acceleration is constant at g, whatever its displacement, and it is not directed towards a central equilibrium point, so the motion is periodic but not simple harmonic. ✓
02
Calculation
[3 marks]

The piston of a small engine moves with simple harmonic motion of amplitude 45 mm while the crankshaft turns at 3000 revolutions per minute. Calculate the angular frequency of the motion, the maximum speed of the piston, and its maximum acceleration.

  • f = 3000 / 60 = 50 Hz, so ω = 2πf = 2π × 50 = 314 rad s⁻¹ (3.1 × 10²). ✓
  • v₀ = ωx₀ = 314 × 0.045 = 14 m s⁻¹. ✓
  • a₀ = ω²x₀ = 314² × 0.045 = 4.4 × 10³ m s⁻², about 450 times g, which is why engine parts must be so strong. ✓
03
Calculation
[3 marks]

A buoy bobs up and down on the sea with simple harmonic motion of amplitude 0.60 m and period 4.0 s. At time t = 0 it passes through its equilibrium position moving upwards. Calculate its displacement and its velocity at t = 0.50 s.

  • ω = 2π / T = 2π / 4.0 = 1.57 rad s⁻¹, and with x = 0 at t = 0 the displacement is x = x₀ sin ωt. ✓
  • x = 0.60 × sin(1.57 × 0.50) = 0.60 × sin(0.785 rad) = 0.60 × 0.707 = 0.42 m above equilibrium. ✓
  • v = ωx₀ cos ωt = 1.57 × 0.60 × 0.707 = 0.67 m s⁻¹ upwards; the same value follows from v = ω√(x₀² − x²) = 1.57 × √(0.36 − 0.18). ✓
04
Analysis
[3 marks]

The displacement-time graph of an object in simple harmonic motion is a sine curve. Describe how the velocity-time graph and the acceleration-time graph of the same motion are related to it, and state the displacement at which the velocity and the acceleration each have their greatest magnitude.

  • Velocity is the gradient of the displacement-time graph, so it is a cosine curve of the same period, a quarter of a cycle (π/2) ahead of the displacement; its magnitude is greatest, ωx₀, when the displacement is zero. ✓
  • Acceleration is the gradient of the velocity-time graph, giving a negative sine curve: it is exactly in antiphase with the displacement, as a = −ω²x requires. ✓
  • The acceleration has its greatest magnitude, ω²x₀, at the extremes x = ±x₀, where the velocity is zero. ✓
05
Calculation
[3 marks]

An object oscillates with simple harmonic motion about a fixed point. Its speed is 0.24 m s⁻¹ when its displacement is 30 mm and 0.18 m s⁻¹ when its displacement is 40 mm. Determine the amplitude and the angular frequency of the motion.

  • v² = ω²(x₀² − x²) at both points: 0.24² = ω²(x₀² − 0.030²) and 0.18² = ω²(x₀² − 0.040²). ✓
  • Subtracting, 0.0576 − 0.0324 = ω²(0.0016 − 0.0009), so ω² = 0.0252 / 0.0007 = 36 and ω = 6.0 rad s⁻¹. ✓
  • Then x₀² = 0.0576 / 36 + 0.0009 = 0.0025, so x₀ = 0.050 m (50 mm); as a check, 6.0 × √(0.0025 − 0.0016) = 0.18 m s⁻¹. ✓
06
Calculation
[4 marks]

A loudspeaker cone vibrates with simple harmonic motion at a frequency of 250 Hz and an amplitude of 0.80 mm.

(a) Calculate the angular frequency of the motion. (b) Calculate the maximum acceleration of the cone. (c) Determine the displacement of the cone at the instants when its speed is half its maximum value.

  • (a) ω = 2πf = 2π × 250 = 1.57 × 10³ rad s⁻¹. ✓
  • (b) a₀ = ω²x₀ = (1.57 × 10³)² × 0.80 × 10⁻³ = 2.0 × 10³ m s⁻². ✓
  • (c) ω√(x₀² − x²) = ½ωx₀, so x₀² − x² = ¼x₀² and x² = ¾x₀². ✓
  • x = (√3 / 2) x₀ = 0.866 × 0.80 = 0.69 mm, on either side of the centre. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.