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What quantities describe a progressive wave, and how are they related?

Wave properties: progressive waves, transverse and longitudinal waves, displacement, amplitude, wavelength, period, frequency and phase, the wave equation, intensity, and polarisation.

A focused answer to the OCR H556 content on wave properties, covering progressive waves, transverse and longitudinal waves, the quantities displacement, amplitude, wavelength, period, frequency and phase difference, the wave equation, intensity and its relation to amplitude, and polarisation.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

OCR wants you to describe progressive waves and distinguish transverse from longitudinal, define displacement, amplitude, wavelength, period, frequency and phase difference, apply the wave equation, relate intensity to amplitude, and explain polarisation.

The answer

Progressive, transverse and longitudinal waves

Describing a wave

The wave equation and phase

Intensity and polarisation

Examples in context

Polarising filters in sunglasses and photography cut glare by blocking light oscillating in one plane. Radio and television aerials are oriented to match the polarisation of the broadcast signal. Sonar and medical ultrasound use longitudinal waves, while seismologists distinguish transverse S-waves from longitudinal P-waves to locate earthquakes (S-waves cannot pass through the liquid outer core). The inverse-square fall of intensity with distance explains why a distant source appears fainter.

Try this

Q1. State the wave equation and define each symbol. [2 marks]

  • Cue. v=fλv = f\lambda: wave speed, frequency and wavelength.

Q2. A wave has a period of 4.0 ms4.0\ \text{ms}. Find its frequency. [1 mark]

  • Cue. f=1T=14.0×10−3=250 Hzf = \frac{1}{T} = \frac{1}{4.0 \times 10^{-3}} = 250\ \text{Hz}.

Q3. Explain why sound waves cannot be polarised. [2 marks]

  • Cue. Sound is longitudinal, oscillating along the direction of travel, so there is no perpendicular plane to restrict; only transverse waves can be polarised.

Exam-style practice questions

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

OCR 20184 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 its period.
Show worked answer →

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

Period is the reciprocal of frequency: T=1f=1256=3.9×10−3 sT = \frac{1}{f} = \frac{1}{256} = 3.9 \times 10^{-3}\ \text{s} (3.9 ms).

Markers reward rearranging v=fλv = f\lambda for wavelength (about 1.3 m1.3\ \text{m}) and using T=1fT = \frac{1}{f} for the period (about 3.9 ms3.9\ \text{ms}).

OCR 20213 marksTwo points on a progressive wave of wavelength 0.60 m0.60\ \text{m} are separated by 0.15 m0.15\ \text{m}. Calculate their phase difference in radians and state whether they are in phase or out of phase.
Show worked answer →

Phase difference is the fraction of a wavelength times 2π2\pi: Δϕ=Δxλ×2π=0.150.60×2π=14×2π=π2 rad\Delta\phi = \frac{\Delta x}{\lambda} \times 2\pi = \frac{0.15}{0.60} \times 2\pi = \frac{1}{4} \times 2\pi = \frac{\pi}{2}\ \text{rad}.

A phase difference of π2\frac{\pi}{2} (a quarter of a cycle) means the points are neither in phase (which needs 00 or 2π2\pi) nor in antiphase (which needs π\pi); they are out of phase by a quarter cycle.

Markers reward the fraction of a wavelength, the value π2 rad\frac{\pi}{2}\ \text{rad}, and the statement that they are out of phase.

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