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What is a wave, and how do transverse and longitudinal waves differ?

Transverse and longitudinal waves, amplitude, wavelength, frequency, period and phase, the wave equation, and polarisation.

A focused answer to WJEC A-Level Physics Unit 2 the nature of waves, covering transverse and longitudinal waves, amplitude, wavelength, frequency, period and phase, the wave equation, and polarisation as evidence of transverse waves.

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

WJEC wants you to distinguish transverse from longitudinal waves, define the wave quantities and phase, use the wave equation, and explain polarisation as evidence that a wave is transverse. These definitions and the wave equation are foundational, reappearing in every later wave topic from interference to standing waves, and polarisation is a favourite short-answer question.

The answer

Transverse and longitudinal waves

In both cases the wave transfers energy without transferring matter: the medium oscillates about a fixed point but does not travel with the wave.

Wave quantities and the wave equation

Amplitude is the maximum displacement from equilibrium. Wavelength λ\lambda is the distance between two adjacent points in phase. Frequency ff is the number of cycles per second and period TT the time for one cycle, with f=1Tf = \dfrac{1}{T}. Phase difference describes how far through a cycle one point is relative to another, usually measured in radians, where a full cycle is 2π2\pi.

The wave equation follows from the definitions: in one period TT the wave advances exactly one wavelength λ\lambda, so v=λ/T=fλv = \lambda / T = f\lambda.

Polarisation

That light can be polarised (for example by a Polaroid filter) is direct evidence that light is a transverse wave.

Examples in context

Example 1. Polarised sunglasses
Light reflected off a wet road or water surface is partly horizontally polarised. Polaroid sunglasses are oriented to block this horizontal component, cutting glare while still letting through vertically polarised light from the rest of the scene. This everyday product is a direct application of the transverse nature of light.
Example 2. Seismic waves locating an earthquake
An earthquake produces fast longitudinal P-waves and slower transverse S-waves. Because S-waves are transverse, they cannot pass through the liquid outer core, and the resulting shadow zone helped reveal the Earth's layered structure. The time gap between the P and S arrivals, combined with the wave equation, lets seismologists pinpoint the distance to the epicentre.
Example 3. Tuning a radio
A radio receiver selects one station by matching its circuit to the frequency of the incoming electromagnetic wave. Since v=fλv = f\lambda and all such waves travel at cc, choosing a frequency is the same as choosing a wavelength. The aerial must also be aligned with the plane of polarisation of the transmitted wave to pick up the strongest signal, which is why car aerials are angled the way they are.

Try this

Q1. A wave has frequency 50Hz50\,\text{Hz} and wavelength 6.0m6.0\,\text{m}. Find its speed. [2 marks]

  • Cue. v=fλ=50×6.0=300m s1v = f\lambda = 50\times6.0 = 300\,\text{m s}^{-1}.

Q2. Explain how polarisation shows that light is a transverse wave. [2 marks]

  • Cue. Only transverse waves can be polarised; light can be polarised, so it must be transverse.

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20194 marksA radio station broadcasts at a frequency of 96.5MHz96.5\,\text{MHz}. Calculate the wavelength of the waves, and state and explain whether a metal rod aerial should be vertical or horizontal if the waves are vertically polarised. Take c=3.0×108m s1c = 3.0 \times 10^{8}\,\text{m s}^{-1}.
Show worked answer →

Use the wave equation v=fλv = f\lambda rearranged for wavelength.

λ=cf=3.0×10896.5×106=3.11m\lambda = \dfrac{c}{f} = \dfrac{3.0 \times 10^{8}}{96.5 \times 10^{6}} = 3.11\,\text{m}.

For a vertically polarised wave, the electric field oscillates in the vertical plane. To pick up the maximum signal the aerial rod should be aligned with the oscillating electric field, so it should be vertical.

If the rod were horizontal it would be perpendicular to the field oscillations and pick up little or no signal. Markers reward the wavelength near 3.1m3.1\,\text{m} and aligning the aerial with the plane of polarisation.

WJEC 20213 marksState two differences between a transverse and a longitudinal wave, and give one example of each.
Show worked answer →

First difference: in a transverse wave the particle oscillations are perpendicular to the direction of energy travel, whereas in a longitudinal wave the oscillations are parallel to the direction of travel.

Second difference: transverse waves can be polarised, but longitudinal waves cannot, because polarisation restricts oscillations to one plane and longitudinal oscillations are already along the travel direction.

Examples: a transverse wave such as light (or a wave on a string); a longitudinal wave such as sound. Markers reward two clear distinguishing features and one correct example of each type.

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