Simple harmonic motion: definition and proof

Definition

A particle moves with SHM about a centre O when

ẍ = -ω² x,

where x is the displacement from O and ω is a positive constant. The minus sign means the acceleration is always towards O. The greatest displacement is the amplitude a, and the time for one full oscillation is the period T = 2π/ω.

Standard results

Pearson lets you quote these without proof:

• x = a sin ωt (starting at O) or x = a cos ωt (starting at an end point)
• v² = ω²(a² - x²)
• T = 2π/ω.

The v² result comes from v dv/dx = -ω^2x, which gives ½v² = -½ω^2x2 + c, with v = 0 at x = a.

Proving SHM

1. Draw the particle at a general displacement x from the centre (not at the equilibrium position).

2. Write F = ma with every force in terms of x.

3. Simplify to ẍ = -(constant)x and state the constant as ω².

4. Conclude: "so the motion is SHM". Pearson's examiners want the result written as ẍ = -ω^2x, and warn against using a for acceleration because a is the amplitude.

If you get ẍ = -k(x - c), the motion is SHM about x = c.

Worked example

A particle has speed v and displacement x from O, and moves so that ẍ = -4x. When x = 0 its speed is 10 m/s. Find the amplitude and period.

1. ω = 2.
2. At the centre v = aω, so 10 = 2a and a = 5 m.
3. T = 2π/2 = π s.

Answer: a = 5 m, T = π = 3.14 s

Check yourself

1. Which equation describes simple harmonic motion?
   • ẍ = 9x
   • ẍ = -9x
   • ẍ = -9
   • ẍ = -9x²
   Show answer

   ẍ = -9x. SHM needs acceleration proportional to x and towards the centre.

2. A particle moves so that ẍ = -16x. Find its period, to 3 significant figures.
   Show answer

   1.57 s. ω = 4, T = 2π/4 = 1.57 s.

3. A particle moves on the x-axis so that ẍ = -9(x - 2). About which value of x does it oscillate?
   Show answer

   x = 2. Let X = x - 2; then Ẍ = -9X, which is SHM about X = 0, that is x = 2.

4. A particle moves with SHM with period 3 s. Find ω, to 3 significant figures.
   Show answer

   2.09 rad/s. ω = 2π/T = 2.09 rad/s.

5. A particle moves with SHM with ω = 2 and amplitude 5 m. Find its speed when it is 3 m from the centre.
   Show answer

   8 m/s. v² = 4(25 - 9) = 64, v = 8 m/s.

6. Which of these is NOT a feature of simple harmonic motion?
   • The acceleration is always directed towards the centre
   • The period does not depend on the amplitude
   • The speed is greatest at the end points
   • The acceleration is zero at the centre
   Show answer

   The speed is greatest at the end points. At the end points the particle is momentarily at rest; it is fastest at the centre.

7. You have shown that a particle's equation of motion simplifies to ẍ = -25x. Which final statement earns the conclusion mark?
   • ẍ = -25x, which is of the form ẍ = -ω^2x with ω = 5, so the motion is SHM
   • a = -25x, so it oscillates
   • The particle moves in a circle
   • The particle has constant acceleration of 25
   Show answer

   ẍ = -25x, which is of the form ẍ = -ω^2x with ω = 5, so the motion is SHM. State the form, identify ω and conclude SHM.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M3
• Pearson Edexcel IAL Mechanics M3 (WME03) question paper, June 2025: front-cover instruction on g and accuracy
• Pearson Edexcel IAL Mechanics M3 (WME03) examiners' report, June 2025

Last updated 8 November 2026. Spotted a mistake? Tell us.