Simple harmonic motion

The defining condition

An object moves with simple harmonic motion if its acceleration is proportional to its displacement from a fixed equilibrium point and always directed towards that point:

F = -kx and a = -ω^2x

The minus sign shows the restoring force acts opposite to the displacement. Examples: a mass on a spring, a simple pendulum at small angles, a ball rolling in a curved dish, a floating object bobbing up and down.

The equations

With x = A at t = 0 (released from maximum displacement):

• x = A cos ωt
• v = -Aω sin ωt
• a = -Aω² cos ωt

where A is the amplitude and ω = 2π / T = 2πf (angular frequency, rad/s). Use radians on your calculator.

Maximum values

• Maximum speed v_max = Aω, at the equilibrium position (x = 0).
• Maximum acceleration a_max = Aω², at the extremes (x = ±A), where the speed is zero.
• The period does not depend on the amplitude: SHM is isochronous.

Worked example

A mass oscillates with amplitude 0.040 m and frequency 2.5 Hz. Find its maximum speed and maximum acceleration.

1. ω = 2πf = 2π x 2.5 = 15.7 rad/s
2. v_max = Aω = 0.040 x 15.7 = 0.628 m/s
3. a_max = Aω² = 0.040 x 15.7²

Answer: v_max = 0.63 m/s; a_max = 9.9 m/s²

Check yourself

1. An oscillator has a frequency of 2.5 Hz. Find its angular frequency in rad/s, to 3 s.f.
   Show answer

   15.7 rad/s. ω = 2πf = 2π x 2.5 = 15.7 rad/s.

2. A mass oscillates in SHM with amplitude 0.040 m and frequency 2.5 Hz. Find its maximum speed in m/s, to 3 s.f.
   Show answer

   0.628 m/s. v_max = Aω = 0.040 x 15.71 = 0.628 m/s.

3. For the same oscillation (A = 0.040 m, f = 2.5 Hz), find the maximum acceleration in m/s², to 3 s.f.
   Show answer

   9.87 m/s². a_max = Aω² = 0.040 x 15.71² = 0.040 x 246.7 = 9.87 m/s².

4. An oscillator with amplitude 6.0 cm and period 0.80 s is released from maximum displacement at t = 0. Find its displacement at t = 0.10 s, in cm, to 3 s.f.
   Show answer

   4.24 cm. ω = 2π / 0.80 = 7.854 rad/s; x = 6.0 cos(0.785 rad) = 6.0 x 0.707 = 4.24 cm.

5. An object in SHM has ω = 10 rad/s. Find the size of its acceleration when its displacement is 0.020 m, in m/s².
   Show answer

   2 m/s². a = ω^2x = 100 x 0.020 = 2.0 m/s² (directed towards equilibrium).

6. Which condition defines simple harmonic motion?
   • Constant acceleration towards the equilibrium position
   • Constant speed in a circle
   • Acceleration proportional to displacement and directed towards the equilibrium position
   • Acceleration proportional to velocity
   Show answer

   Acceleration proportional to displacement and directed towards the equilibrium position. That is a = -ω^2x.

7. Where is the speed of an object in SHM greatest?
   • At the equilibrium position
   • At maximum displacement
   • Halfway between equilibrium and maximum displacement
   • It is the same everywhere
   Show answer

   At the equilibrium position. All the energy is kinetic as it passes through equilibrium.

8. Which of these is NOT simple harmonic motion?
   • A mass oscillating on a spring
   • A ball bouncing repeatedly on a hard floor
   • A pendulum swinging through a small angle
   • A floating cork bobbing gently up and down
   Show answer

   A ball bouncing repeatedly on a hard floor. Between bounces the ball's acceleration is a constant g, not proportional to displacement.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Pearson Edexcel IAL Physics Unit 5 (WPH15) question paper, June 2023: list of data, formulae and relationships

Last updated 7 November 2026. Spotted a mistake? Tell us.