Key Idea: Topic 3.1 is about oscillations — anything that swings back and forth through a middle position: a mass on a spring, a pendulum, a floating cork. It ties together four ideas: what makes motion 'simple harmonic' (the rule a = −ω²x), how long one swing takes (the period formulas), what the displacement, velocity and acceleration graphs look like (and their phase relationships), and how the energy keeps swapping between kinetic and potential. It is examined on Paper 1A (quick MCQs — identify SHM from a graph or its condition, ratio problems on the period, where the speed/acceleration peaks) and on Paper 2 (substitute into T = 2π√(m/k) or T = 2π√(l/g), use quarter-cycle timing, and energy steps like setting the maximum KE equal to the total energy).
📐 Key formulas
Four of these are given in the data booklet (C.1) — you choose the right one rather than memorising it. The energy formula is the only one you must remember.
- acceleration of the object (m s⁻²)
- angular frequency — how fast it oscillates (rad s⁻¹)
- displacement from equilibrium (m)
- period — time for one full oscillation (s)
- mass on the spring (kg)
- spring constant — stiffness (N m⁻¹)
- period — time for one full oscillation (s)
- length of the pendulum (m)
- gravitational field strength (m s⁻²)
- period — time for one full oscillation (s)
- frequency — oscillations per second (Hz)
- angular frequency (rad s⁻¹)
- total energy of the oscillation (J)
- spring constant — stiffness (N m⁻¹)
- amplitude — the largest displacement (m)
🧭 Which equation, and when?
The most-tested decision in this topic: what is the question actually asking for — to test for SHM, find a period, or find an energy/speed?
| What the question wants | Equation to reach for | Watch out for |
|---|---|---|
| Is the motion SHM? Find ω from a-against-x data | a = −ω²x | Slope is −ω², so square-root it to get ω |
| Period of a mass on a spring | T = 2π√(m/k) | Gravity does not appear |
| Period of a simple pendulum | T = 2π√(l/g) | The bob's mass does not appear |
| Switch between T, f and ω | T = 1/f = 2π/ω | f = 1 ÷ T (Hz), ω = 2πf |
| Total energy, or maximum speed | Eₜₒₜₐₗ = ½kA² | Not given — and max KE = Eₜₒₜₐₗ |
🌀 The two oscillators — what each period depends on
| Oscillator | Period formula | Depends on | Does NOT depend on |
|---|---|---|---|
| Mass on a spring | T = 2π√(m/k) | mass m, stiffness k | gravity g |
| Simple pendulum | T = 2π√(l/g) | length l, gravity g | the bob's mass |
📊 The three SHM graphs — phase and where each peaks
| Quantity | Biggest at… | Zero at… | Phase vs displacement x |
|---|---|---|---|
| Displacement x | the ends (turning points) | the centre (equilibrium) | — (the reference) |
| Velocity v | the centre (rushing through) | the ends (momentarily at rest) | leads x by ¼ cycle (90°) |
| Acceleration a | the ends (max displacement) | the centre (a = −ω²x = 0) | antiphase (180°) — mirror of x |
⚡ Energy through one swing
| Position | Kinetic energy (KE) | Potential energy (PE) | Total energy |
|---|---|---|---|
| At the centre (x = 0) | maximum (fastest) | zero | constant |
| At the ends (x = ±A) | zero (momentarily still) | maximum | constant |
| In between | part KE | part PE | constant — they just trade |
✍️ IB-style worked examples
The acceleration a of an oscillating mass is measured at several displacements x from equilibrium. The data lie on a straight line through the origin, a = −81x (a in m s⁻², x in m). State whether the mass moves with SHM, and find its angular frequency ω.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
A 0.80 kg mass hangs from a spring of spring constant k = 320 N m⁻¹ and is set oscillating. Calculate the period and the frequency of the oscillation.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
A pendulum oscillates with SHM at a frequency of 0.50 Hz. Find the period, then the time it takes to travel from one extreme of its swing to the equilibrium (centre) position.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
A 0.50 kg mass on a spring of spring constant k = 200 N m⁻¹ oscillates with an amplitude of 0.10 m. Calculate the total energy of the oscillation and the maximum speed of the mass.
🔒 Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
✅ Quick self-check
Tap each card to reveal the answer.
What are the two conditions for SHM? Acceleration proportional to displacement, and always directed back toward equilibrium — together, a = −ω²x.
A pendulum's bob mass is doubled. What happens to the period? Nothing — the mass is not in T = 2π√(l/g). Only the length l and gravity g matter.
A spring system's m/k is made 4 times bigger. New period? ×√4 = ×2 — the period only doubles, because of the square root (T ∝ √(m/k)).
Where is the speed biggest, and where is the acceleration biggest? Speed is biggest at the centre; acceleration is biggest at the ends (a = −ω²x, so a peaks where x peaks).
What is the phase between velocity and displacement? Velocity leads displacement by a quarter-cycle (90°); acceleration is antiphase (180°) to displacement.
How do you find the maximum speed of an oscillator? Set the maximum KE equal to the total energy: ½mvₘₐₓ² = ½kA², then solve for vₘₐₓ.
🎯 Highest-yield exam reminders
Exam Tips
- SHM needs BOTH: acceleration proportional to displacement AND directed back to equilibrium — the minus sign in a = −ω²x carries that direction. The slope of an a-against-x line is −ω², so square-root it to get ω.
- Pick the period formula by the system: spring → T = 2π√(m/k) (no gravity); pendulum → T = 2π√(l/g) (no bob mass). Both have a square root, so multiplying a quantity by 4 only changes T by √4 = 2.
- Switch freely between T, f and ω with T = 1/f = 2π/ω: f = 1 ÷ T (in Hz) and ω = 2πf (in rad s⁻¹).
- A full cycle splits into four equal quarters of T/4: centre → end → centre → other end → centre. End-to-centre (or centre-to-end) is one quarter, end-to-end is half a period.
- Velocity leads displacement by 90° (a quarter-cycle); acceleration is antiphase (180°). Speed peaks at the centre, acceleration at the ends.
- Energy keeps swapping: KE is maximum at the centre, PE at the ends, but the total stays constant at Eₜₒₜₐₗ = ½kA². Bigger amplitude → more total energy (E ∝ A²).
- Eₜₒₜₐₗ = ½kA² is NOT in the data booklet — remember it. To get the maximum speed, set the maximum KE equal to the total energy (½mvₘₐₓ² = ½kA²).