Simple harmonic motion

A swinging pendulum, a bouncing mass on a spring and a vibrating guitar string all oscillate back and forth about a middle position. When the acceleration is proportional to the displacement and always points back to the middle, the motion is simple harmonic, and its period does not depend on the size of the swing.

Key facts

  • Period T = 1 / f; angular frequency ω = 2πf = 2π / T.
  • SHM: a = −ω²x (acceleration ∝ displacement, towards equilibrium).
  • x = A cos ωt (released from the end); v_max = ωA at the centre; a_max = ω²A at the ends.
  • Speed at displacement x: v = ω√(A² − x²).
  • Mass on a spring: T = 2π√(m/k). Simple pendulum (small swings): T = 2π√(L/g).
  • Total energy = ½kA² = ½mω²A². Damping removes energy; resonance happens when the driving frequency equals the natural frequency.

Worked examples

Example 1

A pendulum is 0.80 m long. Find its period. (g = 9.8 m/s²)

  1. One period is a full swing: out to one side, across, and back.
  2. For small swings the period depends only on the length and g. T = 2π√(0.80 ÷ 9.8) = 1.8 s

Answer: 1.8 s

Example 2

A 0.50 kg mass on a spring (k = 200 N/m) oscillates with amplitude 4.0 cm. Find the period and the maximum speed.

  1. Period of a mass on a spring: T = 2π√(0.50 ÷ 200) = 0.314 s
  2. Angular frequency: ω = √(k/m) = √400 = 20 rad/s
  3. Fastest at the centre: v_max = ωA = 20 × 0.040 = 0.80 m/s

Answer: 0.314 s; 0.8 m/s

Example 3

An object in SHM has amplitude 10 cm and frequency 2.0 Hz. Find its speed when it is 6.0 cm from the centre.

  1. Angular frequency from the frequency. ω = 2πf = 2π × 2.0 = 12.57 rad/s
  2. Mark the object 6.0 cm from the centre.
  3. Use v = ω√(A² − x²). v = 12.57 × √(0.10² − 0.060²) = 12.57 × 0.080 = 1.01 m/s

Answer: 1.01 m/s