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Free resources/Edexcel IAL Physics revision/Oscillations

Edexcel IAL Physics notes · A2 — Mechanics & fields

Oscillations

Simple harmonic motion occurs when acceleration is proportional to displacement and directed towards equilibrium. The restoring mechanism determines the period; damping and driving determine how the motion evolves.

By Dr Desouky Physics AcademyUpdated 25 September 202610 min reading + practice

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In this lesson
  1. 01The key idea
  2. 02Test the acceleration–displacement relationship
  3. 03Relate the restoring force to the period
  4. 04Explain energy transfer during driven motion
  5. 05Key definitions
  6. 06Equations & units
  7. 07Worked examples
  8. 08Practice & solutions
  9. 09Common mistakes
  10. 10Questions, explained
  11. 11Learning checklist
  12. 12Sources & citation

The key idea

Simple harmonic motion: the definition is an equation, and the examiner wants it with the minus sign explained. Everything else — energy interchange, damping, resonance — follows from that one line.

Understand it · 01

Test the acceleration–displacement relationship↗

An oscillation is repeated motion around an equilibrium position. Simple harmonic motion is a special case with a = −ω²x, where x is displacement from equilibrium and ω is constant. The minus sign specifies that acceleration is always directed back towards equilibrium. It does not say that velocity must be negative. At the extreme positions, speed is zero but acceleration magnitude is greatest; at equilibrium, acceleration is zero but speed is greatest.

The motion can be described by sine or cosine depending on the chosen time origin. Angular frequency is 2π/T and has units rad s⁻¹, while ordinary frequency is measured in hertz. A graph of acceleration against displacement is a straight line through the origin with negative gradient −ω². This provides a diagnostic test for SHM; simply observing a regular repeating motion is not enough to establish the required proportionality.

Understand it · 02

Relate the restoring force to the period↗

For an ideal mass–spring oscillator, the restoring force relative to equilibrium is −kx, giving ω² = k/m and T = 2π√(m/k). The result assumes linear spring behaviour and negligible damping; the spring’s own effective moving mass may need consideration in a practical experiment. For a simple pendulum, T = 2π√(L/g) is a small-angle approximation with length measured to the bob’s centre of mass, not just to the end of the string.

In an undamped ideal oscillator, energy continually exchanges between kinetic and potential forms while the total remains constant. For the spring system, total oscillation energy is ½kA² and the maximum speed is ωA. Doubling amplitude therefore doubles maximum speed and quadruples energy without changing the ideal period. Real springs or a large-angle pendulum can depart from these amplitude-independent period predictions when the restoring force is no longer accurately proportional to displacement.

Three positions of a spring oscillator show inward acceleration at both extremes and maximum speed at equilibrium.
Figure 2. Speed and acceleration reach their maxima at different positions.Open full-size SVG diagram ↗

Pause & explain

Where is acceleration greatest in magnitude during SHM?

Check your reasoning

At the two extreme displacements, where |x| = A. It is zero at equilibrium.

Understand it · 03

Explain energy transfer during driven motion↗

Damping transfers energy from an oscillator to its surroundings, usually reducing its amplitude with time. In forced oscillations, an external periodic force supplies energy. The system eventually oscillates at the driving frequency, and the amplitude depends on how efficiently the driver transfers energy relative to the losses. Resonance produces a large response near the natural frequency when damping is light. Greater damping reduces and broadens the resonance peak rather than making all driving frequencies equally effective.

When measuring a pendulum period, time several complete oscillations and divide by their number. Start and stop at the same reference position and in the same direction to avoid counting half-periods. Keep the amplitude small when testing the simple-pendulum equation and measure several lengths. Plot T² against L, for which the predicted gradient is 4π²/g. Timing repeats reduces random uncertainty; consistently measuring the wrong pendulum length introduces a systematic error that repetition cannot remove.

Pause & explain

Why does damping reduce a resonance peak?

Check your reasoning

More energy is dissipated each cycle, so the steady driving force cannot sustain as large an amplitude.

Key definitions

Simple harmonic motion
Motion in which the acceleration is proportional to the displacement from a fixed point and always directed towards that point: a = −ω²x.
Damping
The removal of energy from an oscillating system by resistive forces, causing the amplitude to decrease over time.
Resonance
A large response to periodic driving near a system's natural frequency. Light damping produces a high, narrow amplitude peak; damping can shift the amplitude maximum slightly below the undamped natural frequency.

Equations, units & conditions

Defining equation

a = −ω² x

Displacement

x = x₀ sin(ωt)

Velocity

v = ±ω √(x₀² − x²)

Maximum speed

v_max = ω x₀

Open the Edexcel IAL Physics formula sheet →

Worked examples, step by step

Example 1

A mass oscillates in SHM with amplitude 5.0 cm and period 2.0 s. Find its maximum speed and its speed at x = 3.0 cm.

Follow the method

  1. ω = 2π/T = π rad/s.
  2. v_max = ωx₀ = π × 0.050 = 0.157 m/s.
  3. At x = 0.030 m: v = ω√(x₀² − x²) = π√(0.0025 − 0.0009) = π × 0.040.

v_max ≈ 0.16 m/s; v ≈ 0.13 m/s at x = 3.0 cm

Practice questions with worked solutions

Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.

  1. Build · 3 suggested marks

    1. Spring period and maximum speed

    A 0.50 kg mass oscillates on an ideal spring of stiffness 80 N m⁻¹ with amplitude 0.040 m. Find its period and maximum speed.

    Show solution and self-check: question 1
    1. ω = √(k/m) = √160 = 12.65 rad s⁻¹.
    2. T = 2π/ω = 0.4967 s.
    3. vmax = ωA = 12.65 × 0.040.

    Period 0.497 s; maximum speed 0.506 m s⁻¹.

    Self-check · one mark per point

    • Calculate angular frequency from k/m.
    • Find period using 2π/ω.
    • Use vmax = ωA.
  2. Build · 3 suggested marks

    2. Acceleration from displacement

    An SHM oscillator has period 0.80 s. Find its acceleration when x = +0.025 m.

    Show solution and self-check: question 2
    1. ω = 2π/0.80 = 7.854 rad s⁻¹.
    2. a = −ω²x = −7.854² × 0.025.

    −1.54 m s⁻², towards equilibrium.

    Self-check · one mark per point

    • Convert period to angular frequency.
    • Use a = −ω²x.
    • Give the negative sign and its physical meaning.

Common mistakes & how to avoid them

  • Defining SHM without both conditions: acceleration proportional to displacement AND directed towards the equilibrium position (the meaning of the minus sign). One without the other loses the mark.
  • Maximum speed placed at the extremes. Speed is maximum at the centre (x = 0); acceleration is maximum at the extremes (x = ±x₀). They are out of phase by 90°.
  • Claiming damping changes the frequency dramatically. Light damping reduces amplitude steadily but leaves the period almost unchanged — and shifts the resonance peak only slightly (to a lower frequency, with a lower, broader peak).
  • Explain resonance by relating the driving frequency to the natural frequency and balancing energy supplied with energy dissipated. For light damping the large amplitude peak occurs close to the natural frequency; stronger damping lowers and broadens it.

Understand the why

Questions, explained

Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.

  1. What does the minus sign in a = −ω²x mean?
  2. Where are speed and acceleration greatest in SHM?
  3. What is resonance and when is it a problem?
  4. How do I find the SHM period from an acceleration–displacement graph?
  5. Does a larger amplitude change period, speed or energy?
  6. How can I improve a pendulum-period experiment?
Browse all physics questions →

What does the minus sign in a = −ω²x mean?

The minus sign in a = −ω²x means acceleration points opposite to displacement from equilibrium. It is a restoring acceleration: positive displacement gives negative acceleration, and negative displacement gives positive acceleration.

The sign does not specify the direction of velocity. At a positive displacement, the oscillator may be travelling farther outward or returning inward, but its acceleration still points back towards equilibrium. Both proportionality and direction are required for SHM; a restoring force alone may be nonlinear and produce another type of oscillation. Plotting acceleration against displacement gives a straight line through the origin with negative gradient. The slope determines ω², allowing period to be inferred without directly timing every oscillation. Label the equilibrium position carefully because displacement must be measured from that point.

  • Read the supporting explanation ↑

Where are speed and acceleration greatest in SHM?

In SHM, speed is greatest at equilibrium and zero at the two extremes. Acceleration magnitude is greatest at the extremes and zero at equilibrium. Energy alternates between kinetic and potential forms as the oscillator moves.

At an extreme, the object reverses direction: its velocity is momentarily zero, but the restoring force is largest. At equilibrium, restoring force and acceleration vanish instantaneously, yet the object passes through at its greatest speed because kinetic energy is maximum. These observations do not conflict; velocity describes motion now, while acceleration describes how velocity changes. For a mass on a vertical spring, use displacement from the loaded equilibrium, not extension from the spring’s natural length. The weight shifts equilibrium but does not change the ideal linear oscillation equation about it.

  • Read the supporting explanation ↑

What is resonance and when is it a problem?

Resonance is a large response to periodic driving near a system’s natural frequency. With light damping, supplied energy can sustain a large oscillation amplitude; stronger damping lowers and broadens the response peak and can shift its position.

The driving frequency determines the steady forced oscillation frequency, while the amplitude depends on the relationship between driving and natural frequencies. Energy does not accumulate without limit in a realistic steady state because damping removes it. A swing can be driven effectively with appropriately timed pushes; structures may need damping to avoid damaging motion. Distinguish the natural frequency of the undriven system from the exact frequency of maximum displacement amplitude in a damped system. The simple equality is a useful light-damping approximation, rather than an exact description of every resonance curve.

Amplitude–frequency curves show a tall narrow light-damping peak and a lower broader heavily damped response near the natural frequency.
More damping reduces the peak amplitude and broadens the response.Open full-size SVG diagram ↗
  • Read the supporting explanation ↑

How do I find the SHM period from an acceleration–displacement graph?

The gradient of an acceleration–displacement graph for SHM is −ω². Take the square root of the gradient’s magnitude to find angular frequency, then calculate period from T = 2π/ω.

The graph must have acceleration vertically and displacement horizontally for this direct interpretation. A positive gradient would indicate acceleration away from the origin rather than a stable restoring motion under that sign convention. Check displacement units: centimetres on the axis must be converted before reporting a slope in s⁻². A straight line through the origin supports the proportional restoring model, while curvature can show that the motion is outside the linear range. The period inferred this way describes the modelled motion, so compare it with a direct timing measurement as a useful consistency check.

Period from a graph

An acceleration–displacement graph has gradient −64 s⁻². Find the period.

  1. −ω² = −64, so ω = 8.0 rad s⁻¹.
  2. T = 2π/8.0.

0.785 s.

  • Read the supporting explanation ↑

Does a larger amplitude change period, speed or energy?

For an ideal linear mass–spring oscillator, changing amplitude does not change period, but maximum speed is proportional to amplitude and total energy is proportional to amplitude squared. Real oscillators can depart from amplitude independence outside their model limits.

The period depends on m and k through 2π√(m/k). A larger amplitude means a larger restoring force and greater acceleration, which allow the object to cover the longer path in the same ideal time. That reasoning fails if spring stiffness changes with extension or damping is significant. For a pendulum, the familiar amplitude-independent period is a small-angle approximation; large swings take longer. Keep these assumptions explicit when comparing measurements, because an observed period change may reveal a model limitation rather than a mistake in timing.

Double the amplitude

An ideal spring oscillator’s amplitude rises from 0.020 m to 0.040 m with mass and spring unchanged. Compare period, maximum speed and energy.

  1. Amplitude doubles.
  2. T depends only on m and k, so it is unchanged.
  3. vmax = ωA doubles; total energy ½kA² increases by 2².

Period unchanged; maximum speed doubles; energy quadruples.

  • Read the supporting explanation ↑

How can I improve a pendulum-period experiment?

For a simple pendulum, measure length from the suspension point to the bob’s centre of mass and time several complete oscillations. Keep the angle small if testing T = 2π√(L/g), and use a consistent reference crossing.

Starting and stopping on opposite-direction crossings can introduce a half-period counting error. Timing many cycles reduces the relative contribution of reaction-time uncertainty, but repeat trials are still useful for estimating scatter. The model treats the bob as a point mass and the string as light and effectively inextensible. Measuring only the string length underestimates the pendulum length if the bob radius is appreciable. Plot T² against L for a linear test, and consider whether a nonzero intercept could come from a consistent length offset rather than forcing every line through the origin.

  • Read the supporting explanation ↑

Your learning checklist

Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.

  • Define simple harmonic motion: a = −ω²x
  • Use x = x₀ sin ωt, v = ±ω√(x₀² − x²), and v_max = ωx₀
  • Describe the interchange between kinetic and potential energy in SHM
  • Describe light, critical and heavy damping
  • Describe forced oscillations and resonance; sketch amplitude-frequency curves
Check which exam board you study

Sources & how to cite these notes

Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.

  • Pearson Edexcel International AS/A Level Physics — specification

Reference this page

Dr Desouky Physics Academy. Edexcel IAL Physics: Oscillations. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/oscillations

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