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Notes/Physics/Paper 4/Oscillations
CAIEA Level9702§17.1–17.3

Oscillations

Simple harmonic motion — equations and graphs, energy in SHM, damping, forced oscillations and resonance.

150 min read 8 sub-topics
170
question parts
2021–2025 · 37 papers
9 marks
per paper
≈ 9% of the paper
2.0/3
avg difficulty
moderate
#3
most examined
of 16 topics by marks

Oscillations are the physics of second chances: pull any system away from a stable equilibrium and, when you let go, it comes back — overshoots — and repeats. The guitar string, the suspension of a car, the quartz crystal timing a watch, the atoms of every solid: one mathematical description governs them all, and it is short enough to memorise. This note builds that description from scratch. The single most important object is the defining condition of simple harmonic motion — acceleration proportional to displacement and pointed back towards the fixed point, a=−ω2xa = -\omega^2 x — and almost every exam question is a direct raid on what follows from it.

The bank agrees on the importance: 339 marks across 2021–2025 rank Oscillations 3rd of the 16 A2 topics, carried by 170 question parts in 37 paper sittings — roughly 9 marks on every Paper 4. The yearly haul reads 75, 65, 52, 63, 84, with a record 84 in 2025. Yet the mean difficulty of 2.01 is among the gentlest of the A2 tier, and the reason is structural: nearly every mark hangs off one small family of equations (T=2π/ωT = 2\pi/\omega, a=−ω2xa = -\omega^2 x, v=± ωx02−x2v = \pm\,\omega\sqrt{x_0^2 - x^2}, E=12mω2x02E = \tfrac{1}{2}m\omega^2 x_0^2) plus a set of word-perfect definitions the schemes reprint year after year. Learn the family once and the topic pays rent for the rest of the course.

The route through is: §01 the vocabulary — amplitude, period, frequency, angular frequency, phase difference — and the defining condition in the examiner's own words — §02 where x=x0sin⁡ωtx = x_0 \sin\omega t comes from, by projecting uniform circular motion — §03 acceleration, a=−ω2xa = -\omega^2 x, and the straight-line aa–xx graph that proves SHM — §04 velocity, v=± ωx02−x2v = \pm\,\omega\sqrt{x_0^2 - x^2} — §05 reading and sketching the displacement, velocity and acceleration graphs like an examiner — §06 the energy ledger E=12mω2x02E = \tfrac{1}{2}m\omega^2 x_0^2 and the kinetic–potential interchange — §07 damping, from light through critical to heavy — and §08 forced oscillations and resonance, where the whole story peaks.

Before you start you should be able to

  • Use the A2 Motion in a Circle results fluently: a point moving on a circle of radius rr at angular velocity ω\omega has speed v=rωv = r\omega and acceleration a=rω2a = r\omega^2 directed towards the centre

  • Read gradients and values off motion graphs quickly — this topic lives on displacement–time, velocity–time and acceleration–time axes

  • Work confidently in radians: sin⁡ωt\sin\omega t with ω\omega in rad s⁻¹, and angles advancing steadily as θ=ωt\theta = \omega t

  • Recall from AS Deformation that a spring stretched or compressed by xx exerts a restoring force F=kxF = kx towards equilibrium — the standard physical example of SHM

  • Handle powers of ten without ceremony: amplitudes arrive in mm and cm, periods in ms and μs, so unit conversion happens before any substitution

By the end of this page you can

  • Understand and use the terms displacement, amplitude, period, frequency, angular frequency and phase difference in the context of oscillations, and express the period in terms of both frequency and angular frequency

  • Understand that simple harmonic motion occurs when acceleration is proportional to displacement from a fixed point and in the opposite direction

  • Use a=−ω2xa = -\omega^2 x and recall and use, as a solution to this equation, x=x0sin⁡ωtx = x_0 \sin\omega t

  • Use the equations v=v0cos⁡ωtv = v_0 \cos\omega t and v=± ωx02−x2v = \pm\,\omega\sqrt{x_0^2 - x^2}

  • Analyse and interpret graphical representations of the variations of displacement, velocity and acceleration for simple harmonic motion

  • Describe the interchange between kinetic and potential energy during simple harmonic motion, and recall and use E=12mω2x02E = \tfrac{1}{2}m\omega^2 x_0^2 for the total energy

  • Understand that a resistive force acting on an oscillating system causes damping; understand and use the terms light, critical and heavy damping and sketch displacement–time graphs illustrating them

  • Understand that resonance involves a maximum amplitude of oscillations and occurs when an oscillating system is forced to oscillate at its natural frequency

01

The language of oscillations

Syllabus requirement · §17.1.1

“

understand and use the terms displacement, amplitude, period, frequency, angular frequency and phase difference in the context of oscillations, and express the period in terms of both frequency and angular frequency; understand that simple harmonic motion occurs when acceleration is proportional to displacement from a fixed point and in the opposite direction

”

Why wobbles get their own chapter

Nudge anything away from a stable equilibrium — pull a pendulum bob sideways, compress a car's suspension, pluck a guitar string — and the restoring effect of the equilibrium takes over: the system returns, overshoots, returns again. One repeated journey there and back is one oscillation (or one complete vibration), and an astonishing range of systems perform oscillations that are geometrically identical: same-shaped graph, same equations, different labels on the axes. That shared shape is worth a chapter of its own, because the examiner can test it with a mass on a spring one year and a vibrating crystal the next and mark both with the same scheme.

The workhorse picture for this section is a cart of mass mm sitting on a frictionless surface between two springs, drawn below. Its equilibrium position is where it rests when undisturbed — the springs neither stretched nor compressed, net force zero. Every quantity defined below is measured from that position, and forgetting so is the topic's most common unit-level error.

equilibriummF = 0 hereF pushes backF pulls back+x−xwhy it comes backsprings stretch or squash⇒ restoring force towardsthe equilibrium positionHooke's law: F ∝ x, soa ∝ −xeither side — the sign flips,the aim never does

The workhorse oscillator: a cart of mass m between two springs. Displace it either side of equilibrium and the spring force always aims back at the centre while growing with distance — F ∝ −x, hence acceleration a ∝ −x. That double condition is simple harmonic motion, defined in full later this section.

The six words the examiner owns

Displacement, xx: the distance of the mass from its equilibrium position, including direction — positive on one side, negative on the other. Displacement is not distance travelled: a mass mid-swing on the left-hand side has negative displacement even though it may have travelled far. It changes continuously throughout the oscillation.

Amplitude, x0x_0: the maximum displacement — the size of the excursion to either side, always written as a positive quantity. If the cart swings between +3 cm+3\ \text{cm} and −3 cm-3\ \text{cm}, the amplitude is 3 cm3\ \text{cm}, not 6 cm: amplitude is measured from the middle, never peak-to-peak.

Period, TT: the time for one complete oscillation — from some state, around the whole journey, back to the same state moving the same way.

Frequency, ff: the number of complete oscillations per unit time. Frequency and period are reciprocals: 50 oscillations per second means each one lasts 150 s\tfrac{1}{50}\ \text{s}.

Angular frequency, ω\omega: the oscillation rewritten in radians. Imagine the oscillation as the shadow of steady circular motion (§02 makes this literal): ω\omega is the rate at which the corresponding angle advances, ω=2πf\omega = 2\pi f, in rad s⁻¹. It earns its own name because x=x0sin⁡ωtx = x_0\sin\omega t is unreadable without it.

Phase difference: how far two oscillations of the same frequency are out of step, expressed as a fraction of a cycle. If two carts cross the centre Δt\Delta t apart, their phase difference is φ=2πΔt/T\varphi = 2\pi\Delta t/T, in radians — one full period of delay being a full 2π2\pi of phase.

quantity

symbol

definition

unit

displacement

x

distance from the equilibrium position, signed

m

amplitude

x₀

maximum displacement from equilibrium

m

period

T

time for one complete oscillation

s

frequency

f

number of oscillations per unit time

Hz

angular frequency

ω

rate of advance of the oscillation, ω = 2πf

rad s⁻¹

phase difference

φ

out-of-step fraction of a cycle, φ = 2πΔt/T

rad

Six terms, examined verbatim. 'Per unit time' does the work in the frequency definition — 'per second' loses the mark when the unit of time is not seconds.

One sentence to carry

Displacement measures distance from equilibrium; amplitude is its maximum; and the three time descriptors lock together as T = 1/f = 2π/ω.

T=1f=2πωT = \frac{1}{f} = \frac{2\pi}{\omega}

Expresses the period in terms of BOTH frequency and angular frequency — exactly what the syllabus outcome demands. Given any one of T, f, ω you own the other two.

φ=2π ΔtT\varphi = \frac{2\pi\,\Delta t}{T}

Phase difference between two identical oscillations whose matching events are delayed by Δt. Answer in radians — the syllabus never asks for degrees here.

What makes an oscillation simple harmonic

Now the condition that separates the special case the whole course studies from wobbles in general. Displace the cart to the right: both springs now push or pull it back towards equilibrium — a restoring force, always aimed at the fixed central point. Close to equilibrium, this force grows in proportion to how far you have displaced the cart (Hooke's law behaviour from AS Deformation). Force gives mass acceleration, so the acceleration obeys the same pattern: proportional to the displacement, and pointing the opposite way.

That double condition defines simple harmonic motion, and the definition is asked verbatim for two marks, paid clause by clause:

acceleration is (directly) proportional to the displacement from a fixed point (B1);
the acceleration acts in the opposite direction to the displacement — i.e. towards the fixed point (B1).

Both clauses are load-bearing. Drop "proportional" and any wobble qualifies; drop "opposite" and you have described a runaway, not an oscillation — a force pointing along the displacement would fling the cart away, not bring it home. In symbols the pair collapses to one line, a∝−xa \propto -x, which §03 upgrades to a=−ω2xa = -\omega^2 x. Everything else in this note is consequences of that minus sign.

Invented demo — one oscillation, four descriptors

A child on a swing completes 45 full oscillations in 60 s, swinging between points 0.25 m either side of the vertical.
Determine the amplitude, the period, the frequency and the angular frequency of the swing, and verify that your values satisfy T=2π/ωT = 2\pi/\omega.

Show full working
  1. 1

    Amplitude first: the swing travels 0.25 m to either side of the vertical (the equilibrium position), so

    x0=0.25 mx_0 = 0.25\ \text{m}

    Measured from the MIDDLE, not peak-to-peak: the total travel is 0.50 m, but amplitude counts one arm only.

  2. 2

    Period: 45 complete oscillations take 60 s, so one oscillation takes

    T=6045=1.33 sT = \frac{60}{45} = 1.33\ \text{s}

    'Complete oscillation' means back to the same state moving the same way — one full there-and-back.

  3. 3

    Frequency: the reciprocal,

    f=1T=4560=0.75 Hzf = \frac{1}{T} = \frac{45}{60} = 0.75\ \text{Hz}

    f = 1/T run forwards: 45 oscillations divided by 60 s. Same numbers, either direction.

  4. 4

    Angular frequency:

    ω=2πf=2π×0.75=4.7 rad s−1\omega = 2\pi f = 2\pi \times 0.75 = 4.7\ \text{rad s}^{-1}

    Radians per second: each oscillation carries the phase through 2π rad, 0.75 times per second.

  5. 5

    Check: 2π/ω=2π/4.712=1.33 s=T2\pi/\omega = 2\pi/4.712 = 1.33\ \text{s} = T ✓ — the three descriptors agree.

    Ten seconds of checking that locks in which formula converts which pair — the reflex every later section assumes.

Answer

x₀ = 0.25 m; T = 1.33 s; f = 0.75 Hz; ω = 4.7 rad s⁻¹, and 2π/ω returns the period.

Whatever else an oscillation question hands you, immediately convert to all three of T, f and ω — later parts almost always want ω.

Common mistakes

  • Reporting the amplitude as the peak-to-peak travel (6 cm for a swing between ±3 cm).

    Amplitude is the maximum DISPLACEMENT from equilibrium: 3 cm.

    Amplitude is measured from the middle of the motion. Peak-to-peak is twice it — a guaranteed lost mark when the value feeds a later part.

  • Quoting ω in hertz, or f in rad s⁻¹.

    f is in hertz (oscillations per second); ω = 2πf is in rad s⁻¹.

    They differ by a factor of 2π ≈ 6.28 — large enough to wreck every downstream calculation while looking superficially plausible.

  • Defining SHM as "acceleration is proportional to the force".

    Acceleration is proportional to the DISPLACEMENT from a fixed point and opposite in direction to it.

    The scheme pays for displacement language. Restating Newton's second law instead of the defining condition earns nothing.

  • Writing only half the definition: "acceleration is proportional to displacement".

    Both clauses: proportional AND in the opposite direction (towards the fixed point).

    Two clauses, two B1s. Without the direction clause the statement describes a system that accelerates away and never oscillates.

Your turn

A one-mark definition the bank reprints, the full two-mark SHM definition, and a phase-difference calculation.

  1. 19702/44 O/N 2025 Q4(a)1 mark

    Define frequency of an oscillation.

    Stuck? Show hint

    It is a count per something.

    Show solution
    1. 1

      The frequency is the number of oscillations per unit time (B1).

      One clause, one mark. 'Number of oscillations per second' over-specifies — the unit of time need not be the second until the unit of f is attached.

    Answer

    The number of oscillations (complete cycles) per unit time.

  2. 22 marks

    State what is meant by simple harmonic motion.

    Stuck? Show hint

    Two clauses: what the acceleration is proportional to, and which way it points.

    Show solution
    1. 1

      Clause one: the acceleration is (directly) proportional to the displacement from a fixed point (B1);

      'From a fixed point' locates the measurement — displacements are measured from equilibrium, nowhere else.

    2. 2

      Clause two: the acceleration is in the opposite direction to the displacement — i.e. directed towards the fixed point (B1).

      The direction clause is what turns proportionality into an oscillation. Omit it and the answer scores one of the two marks.

    Answer

    Motion in which acceleration is directly proportional to displacement from a fixed point and is directed opposite to the displacement (towards the fixed point).

  3. 39702/42 M/J 2022 Q4(b)(iii)2 marks

    A heavy pendulum and a light pendulum are suspended from the same piece of string, which is secured at each end to fixed points. Both pendulums have the same natural frequency. The heavy pendulum is set oscillating and, as it oscillates, it causes the light pendulum to oscillate. Fig. 4.2 shows the variation with time tt of the displacements of the two pendulums for three oscillations. The displacement xx of the light pendulum is given by x=0.25sin⁡5.0πtx = 0.25\sin 5.0\pi t (with xx in centimetres and tt in seconds), from which an earlier part establishes the period T=0.40 sT = 0.40\ \text{s}.
    Determine the magnitude of the phase difference ϕ\phi between the oscillations of the light and heavy pendulums. Give a unit with your answer.

    Fig. 4.2 from the question paper: displacement against time for the heavy pendulum and the light pendulum over three oscillations. Matching features of the two traces are offset by a quarter of a period.

    Fig. 4.2 from the question paper: displacement against time for the heavy pendulum and the light pendulum over three oscillations. Matching features of the two traces are offset by a quarter of a period.

    Stuck? Show hint

    Read the time offset Δt between matching points of the two traces, then φ = 2πΔt/T — and remember the unit.

    Show solution
    1. 1

      Read the offset from Fig. 4.2: matching features of the two traces (for example, corresponding upward zero-crossings) are separated by a quarter of a cycle, i.e.

      Δt=0.10 s\Delta t = 0.10\ \text{s}

      Read a gap between LIKE features — same kind of point, same direction of motion — or the offset picks up an extra half-cycle.

    2. 2

      Substitute into the phase rule (C1):

      φ=2π ΔtT=2π×0.100.40\varphi = \frac{2\pi\,\Delta t}{T} = \frac{2\pi \times 0.10}{0.40}

      Δt is the delay between matching events; T converts that delay into a fraction of a full 2π cycle.

    3. 3

      Evaluate (A1):

      φ=π2≈1.6 rad\varphi = \frac{\pi}{2} \approx 1.6\ \text{rad}

      A quarter of a period behind is a quarter of 2π. Quoting '1.6' without its unit forfeits the mark — the question says 'give a unit'.

    4. 4

      Because the oscillations repeat every cycle, the same offset may equally be measured the other way round, Δt=0.30 s\Delta t = 0.30\ \text{s}:

      φ=2π×0.300.40≈4.7 rad\varphi = \frac{2\pi \times 0.30}{0.40} \approx 4.7\ \text{rad}

      — the scheme accepts 1.6 rad or 4.7 rad, provided the unit appears.

      Both descriptions are true of repeating motion; examiners credit either, which is why the scheme lists both answers.

    Answer

    φ = 2π × 0.10/0.40 ≈ 1.6 rad (equivalently 2π × 0.30/0.40 ≈ 4.7 rad), unit rad.

Practise SHM terminology questionsReal past-paper questions · Simple harmonic motion terminology

The rest of this note

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Can you do all of these?

  • Define SHM word-perfectly: acceleration directly proportional to displacement from a fixed point AND in the opposite direction to the displacement

  • Define frequency as the number of oscillations per unit time, and convert both ways along T = 1/f = 2π/ω

  • Derive x = x₀ sin ωt by projecting uniform circular motion, and read amplitude and angular frequency straight off any x(t) expression

  • Recognise SHM from an a–x graph: straight line through the origin (a ∝ x) with negative gradient (a opposite to x); the gradient equals −ω²

  • Compute a(max) = ω²x₀ and place it correctly: maximum at the extremes, zero at the centre

  • Quote v₀ = ωx₀ at the centre and evaluate v = ±ω√(x₀² − x²) at any other displacement

  • State the phase facts: velocity leads displacement by a quarter of a cycle; acceleration is antiphase (π rad out of step) with displacement

  • Read amplitude, period and phase off x–t, v–t and a–t graphs — counting zero crossings without halving gives a factor-of-two trap

  • Describe the KE↔PE interchange in clauses: KE maximum at zero displacement, PE maximum at maximum displacement, sum constant when undamped

  • Recall and use E = ½mω²x₀², including deducing m, ω or x₀ from pairs of graphs

  • Define damping word-perfectly (loss of energy of oscillations due to resistive forces) and sketch light, critical and heavy damping on x–t axes

  • Define resonance word-perfectly (maximum amplitude when driving frequency = natural frequency) and sketch how increased damping lowers, flattens and broadens the resonance curve

  • Convert units before substituting: cm → m, mm → m, μs → s — the arithmetic is easy; the powers of ten are where marks go missing

Now do the questions
170 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes
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