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What are the defining features of a wave, and how do we describe and represent progressive waves?

The nature of waves: transverse and longitudinal progressive waves, the wave quantities and the wave equation, the relationship between phase and path difference, and polarisation.

A focused answer to the Eduqas A-Level Physics Component 3 nature of waves content, covering transverse and longitudinal progressive waves, the wave quantities and the wave equation v = f lambda, the link between phase and path difference, and the polarisation 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

Eduqas wants you to distinguish transverse from longitudinal progressive waves, define the wave quantities (amplitude, wavelength, frequency, period and speed) and use the wave equation v=fλv = f\lambda, relate phase difference to path difference, and explain polarisation as a property of transverse waves only.

The answer

Transverse and longitudinal waves

Wave quantities and the wave equation

Phase and path difference

Polarisation

Examples in context

The wave equation underpins everything from musical acoustics to radio and radar. Polarisation is used in sunglasses and camera filters to cut glare, in LCD screens, in 3D cinema glasses, and in photoelastic stress analysis where polarised light reveals stress concentrations in transparent models. Phase and path difference explain the interference patterns used to test optical surfaces and to read the pits on a CD or DVD.

Try this

Q1. State the difference between a transverse and a longitudinal wave. [2 marks]

  • Cue. In a transverse wave the oscillations are perpendicular to the direction of energy transfer; in a longitudinal wave they are parallel to it.

Q2. A wave has speed 1500 m s−11500\ \text{m s}^{-1} and frequency 30 kHz30\ \text{kHz}. Find its wavelength. [2 marks]

  • Cue. λ=vf=150030 000=0.050 m\lambda = \frac{v}{f} = \frac{1500}{30\,000} = 0.050\ \text{m}.

Q3. State the phase difference, in radians, between two points one wavelength apart on a progressive wave. [1 mark]

  • Cue. 2π2\pi radians (they are in phase).

Exam-style practice questions

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

Eduqas 20183 marksA sound wave of frequency 256 Hz256\ \text{Hz} travels through air at 340 m s−1340\ \text{m s}^{-1}. Calculate its wavelength, and state whether it is transverse or longitudinal.
Show worked answer →

Wavelength from the wave equation v=fλv = f\lambda: λ=vf=340256=1.33 m\lambda = \dfrac{v}{f} = \dfrac{340}{256} = 1.33\ \text{m}.

A sound wave is longitudinal: the air particles oscillate parallel to the direction of energy transfer, forming compressions and rarefactions.

Markers reward λ=vf=1.33 m\lambda = \frac{v}{f} = 1.33\ \text{m} and identifying the sound wave as longitudinal.

Eduqas 20214 marksExplain what is meant by polarisation, state why only transverse waves can be polarised, and give one practical application of polarisation.
Show worked answer →

Polarisation means restricting the oscillations of a transverse wave to a single plane containing the direction of propagation (plane-polarised light).

Only transverse waves can be polarised because their oscillations are perpendicular to the direction of travel, so there is a choice of plane to restrict. In a longitudinal wave the oscillation is along the direction of travel, so there is no transverse plane to select and it cannot be polarised.

An application: polarising sunglasses cut horizontally polarised glare reflected from water or roads; other examples are LCD screens or stress analysis using polarised light. Markers reward the definition, the perpendicular-oscillation reason, and a valid application.

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