The big idea: An object in simple harmonic motion (SHM) swings back and forth — like a mass on a spring or a pendulum.
Three things change as it swings: its displacement x (how far from the centre), its velocity v (how fast), and its acceleration a (how its velocity is changing).
Plotted against time, all three are smooth waves (sinusoids) — but they are shifted relative to one another.
New words — equilibrium, phase, antiphase: Equilibrium = the centre, where the object would rest if it weren't moving (x = 0).
Phase = where you are in the cycle. Two curves are in phase if they peak together, and antiphase if one peaks exactly as the other troughs (half a cycle apart).
The two phase rules to remember: Velocity leads displacement by a quarter-cycle (90°) — v is biggest at the centre, zero at the ends.
Acceleration is antiphase to displacement (180°) — a always points back toward the centre, opposite to x. That's the SHM rule a = -ω²x.
Each full swing takes one period T. Because the curve is symmetric, one cycle splits into four equal quarters, and each quarter takes T/4.
The defining rule for SHM links the acceleration to the displacement, and it is given in the data booklet:
- acceleration of the oscillator (m s⁻²)
- angular frequency (rad s⁻¹)
- displacement from equilibrium (m)
- period — time for one full oscillation (s)
- frequency — oscillations per second (Hz)
- angular frequency (rad s⁻¹)
| Point in the cycle | Displacement x | Velocity v | Acceleration a |
|---|---|---|---|
| At the centre (equilibrium) | zero | maximum | zero |
| At a turning point (the ends) | maximum | zero | maximum |
Quarter-cycle timings: Centre → end takes T/4. End → centre takes another T/4. Centre → opposite end → back to centre is T/2 (half a cycle).
These fractions of T are the heart of most timing questions.
Worked example — angular frequency from the period
A mass on a spring completes one full oscillation every 0.50 s. Find its angular frequency ω.
Solution
- Start with the given formula linking T and ω:
- Rearrange to make ω the subject:
- Put in the period (T = 0.50 s):
- Work it out — keep the unit:
Final answer
ω = 13 rad s⁻¹ (to 2 sig figs).
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How this is tested: These graphs almost always come up as a quick timing or phase question.
- Paper 1A: read a time to reach a point in the cycle off the idea that each quarter takes T/4 (e.g. equilibrium → maximum displacement), or identify the phase between x, v and a. - Paper 2: describe how v and a change as the object moves, using a = -ω²x.
Classic trap: thinking velocity is biggest at the ends — it is biggest at the centre, and zero at the ends.
Split the cycle into quarters: Mark the cycle: centre → end → centre → other end → centre. Each step is T/4.
So equilibrium to maximum displacement is the first quarter = T/4.
IB-style question — time from equilibrium to maximum
A particle moves with simple harmonic motion of period 0.80 s. It starts at the equilibrium position. Find the time it takes to first reach maximum displacement.
Solution
- Equilibrium → maximum displacement is the first quarter of the cycle:
- Put in the period (T = 0.80 s):
- Work it out — keep the unit:
Final answer
time = 0.20 s (one quarter of the 0.80 s period).