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.
[3 marks]
Total for this question: 3
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.
Board and spec code confirmed against AQA 7408 · registry checked 2026-07-11How this checking works
Know the equation
| Symbol | Quantity | Unit |
|---|---|---|
| x | displacement from equilibrium | m |
| A | amplitude | m |
| omega | angular frequency | rad/s |
| t | time | s |
| a | acceleration | m/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
Do the calculation on paper first — then mark it.
Q2
A particle oscillates with frequency 4.0 Hz. At one instant its displacement is +0.035 m. Calculate its acceleration at that instant.
[3 marks]
Total for this question: 3
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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