All equation deep-dives

Simple harmonic motion

Further mechanics
AQA 7408
In the data booklet
ORIGINAL

x = A cos(omega t) and a = -omega2 x

The sign gives the direction: SHM acceleration always points back towards equilibrium.

Verified against AQA 7408 (2026 spec)

Know the equation

SymbolQuantityUnit
xdisplacement from equilibriumm
Aamplitudem
omegaangular frequencyrad/s
ttimes
aaccelerationm/s2

Rearrangements

  • A = x / cos(omega t)
  • omega = 2 pi f = 2 pi / T
  • x = -a / omega2
  • omega = sqrt(-a / x)

Apply it — mark your own working

Work each one out on paper first, then reveal the mark scheme and tick the marks you actually earned. That is exactly how you should mark past papers.

Q1
3 marks

An oscillator has amplitude 0.080 m and frequency 2.5 Hz. At t = 0 it is at maximum positive displacement. Calculate its displacement at t = 0.12 s.

Do the calculation on paper first — then mark it.

Q2
3 marks

A particle oscillates with frequency 4.0 Hz. At one instant its displacement is +0.035 m. Calculate its acceleration at that instant.

Do the calculation on paper first — then mark it.

Where the marks get lost

  • Using frequency f in place of angular frequency omega; convert with omega = 2 pi f first.
  • Dropping the minus sign in a = -omega2 x, which loses the restoring-direction physics.
  • Using degrees mode for cos(omega t); omega t is an angle in radians.

Exam tip: State the sign of your answer. A negative displacement is a position on the negative side; a negative acceleration means the acceleration points in the negative direction.

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