OxfordAQA International AS and A-level Physics: Progressive waves, longitudinal and transverse waves, polarisation (9630)

This guide follows sections 3.5.3 (Progressive waves) and 3.5.4 (Longitudinal and transverse waves) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards. Both sections are shared AS and A-level content. Use it alongside the revision notes and the practice questions. The course hub and checklist cover the rest of the course, and the measurements guide covers units and significant figures.

What this unit covers

Section	What you must be able to do
3.5.3	Describe how the particles of a medium oscillate as a wave passes. Define amplitude, frequency, wavelength, speed, phase and phase difference. Use c = fλ and f = 1/T. Give phase difference as an angle (radians or degrees) or as a fraction of a cycle.
3.5.4	Describe longitudinal and transverse waves, using sound, light and radio (electromagnetic) waves, and a vibrating string, as examples. State the direction of particle (or field) displacement relative to the direction of energy propagation. Know that electromagnetic waves of every kind move at one common speed in a vacuum. Describe the use of ultrasound in medicine. Explain polarisation as evidence for transverse waves, with Polaroid and the alignment of aerials as applications.

The specification says that Malus’s law will not be expected, so this guide does not use it.

What a progressive wave is

A progressive wave carries energy across a region, but the medium itself stays put overall. Each particle of the medium oscillates about its own equilibrium position, and its neighbours copy the motion a little later. A cork on a water wave bobs up and down while the wave moves on.

Electromagnetic waves need no medium. What oscillates is the electric and magnetic field at each point, so the specification speaks of “particles/fields”.

Quantities and definitions

Quantity	Meaning	Unit
Displacement	Distance of a particle from its equilibrium position, in a stated direction	m
Amplitude A	Maximum displacement from the equilibrium position	m
Wavelength λ	Distance between two adjacent points that oscillate in phase, for example crest to crest	m
Period T	Time for one complete oscillation of a particle	s
Frequency f	Number of complete oscillations per second	Hz
Wave speed c	Distance moved by the wave pattern per second	m s⁻¹

The two equations you need are:

• f = 1/T
• c = fλ

The second follows because the pattern moves forward one wavelength in one period, so c = λ/T = fλ.

The source sets the frequency, so it stays the same when a wave enters a new medium. Speed and wavelength change together.

Worked example 1: an ocean swell

A swell has wavelength 48 m and speed 7.5 m s⁻¹. Find its frequency and period.

f = c / λ = 7.5 / 48 = 0.156 Hz
T = 1 / f = 1 / 0.15625 = 6.4 s

A fixed buoy rises and falls once every 6.4 s.

Phase and phase difference

Phase describes which stage of its oscillation a particle has reached. Two particles are in phase when they are at the same stage and moving the same way, so their separation along the wave is a whole number of wavelengths. They are in antiphase when one is at a crest-like position and the other at the matching trough-like position, a phase difference of π rad (180°, half a cycle).

One complete cycle is 360° or 2π rad. The phase difference between two points on a wave is:

phase difference (in cycles) = separation along the wave / λ
phase difference (in radians) = 2π × separation / λ

The same idea works in time. For one particle observed at two instants Δt apart, the fraction of a cycle is Δt/T.

A point further along the direction of travel receives the disturbance later, so it lags behind the point nearer the source.

Worked example 2: phase difference in space and time

A wave has wavelength 0.60 m. Find the phase difference between points 0.15 m apart, and between points 0.50 m apart. Give each as a fraction of a cycle, in degrees and in radians.

0.15 m: 0.15 / 0.60 = 0.25 cycle = 90° = π/2 rad
0.50 m: 0.50 / 0.60 = 0.833 cycle = 300° = 5π/3 rad

Now a source of frequency 40 Hz. Compare the motion of one particle at two instants 5.0 ms apart.

T = 1 / 40 = 0.025 s = 25 ms
5.0 / 25 = 0.20 cycle = 72° = 0.4π rad (1.26 rad)

A difference of 300° is equivalent to 60° the other way round, so say which point leads and which lags if asked.

Reading graphs

There are two different graphs and they are easy to confuse.

Graph	Shows	You read off
Displacement against distance	A snapshot of the whole wave at one instant	Wavelength, amplitude
Displacement against time	The motion of one particle	Period, amplitude

To find the speed you need one of each graph, or a frequency and a wavelength.

Worked example 3: wavelength from a phase experiment

A loudspeaker emits sound of frequency 2.70 kHz. A microphone is moved away from it along a straight line, and its signal is compared with the loudspeaker signal on an oscilloscope. The traces are in phase at distances of 12.3, 24.9, 37.4 and 50.1 cm from the loudspeaker. Find the wave speed.

Four in-phase positions in a row are separated by three wavelengths.

3λ = 50.1 - 12.3 = 37.8 cm
λ = 12.6 cm = 0.126 m
c = fλ = 2700 × 0.126 = 340 m s⁻¹

Measuring across several wavelengths reduces the percentage uncertainty in λ.

Transverse and longitudinal waves

The classification depends on the direction of oscillation compared with the direction in which the energy travels.

	Transverse	Longitudinal
Oscillation direction	At 90° to the direction of energy propagation	Along the direction of energy propagation
Pattern	Crests and troughs	Compressions and rarefactions
Examples	Waves on a string, electromagnetic waves	Sound in air
Can be polarised?	Yes	No

Waves on a string

Flick one end of a stretched string up and down and a pulse travels along it while each piece of string moves up and down, at right angles to the travel. The wave is transverse.

Sound

Air particles oscillate to and fro along the line of travel. Where they are pushed together there is a compression (higher pressure); where they are spread out there is a rarefaction (lower pressure). Sound is carried by particles, so it has no route across empty space.

At the centre of a compression the particles have zero displacement and are moving forwards, in the direction of travel, at their greatest speed.

Electromagnetic waves

An electromagnetic wave is an electric field and a magnetic field oscillating perpendicular to one another, and both perpendicular to the way the energy moves. The wave is transverse. In a vacuum, every electromagnetic wave from radio to gamma has the same speed, 3.00 × 10⁸ m s⁻¹. They differ in frequency and wavelength, and c = fλ links them.

Worked example 4: electromagnetic waves

A radio station broadcasts at 91.5 MHz. Light from a lamp has wavelength 560 nm. Sunlight takes how long to cross 1.50 × 10¹¹ m?

radio:  λ = c / f = 3.00 × 10⁸ / 91.5 × 10⁶ = 3.28 m
light:  f = c / λ = 3.00 × 10⁸ / 560 × 10⁻⁹ = 5.36 × 10¹⁴ Hz
time:   t = d / c = 1.50 × 10¹¹ / 3.00 × 10⁸ = 500 s

Ultrasound in medicine

Ultrasound is sound whose frequency lies beyond human hearing, above about 20 kHz. Medical imaging uses frequencies of a few megahertz. A probe contains a transducer that emits short pulses and also detects the echoes.

Pulse-echo imaging works like this:

1. A short pulse enters the body. A gel between probe and skin removes the air gap, because an air-skin boundary would reflect most of the ultrasound.
2. At each boundary between tissues of different density or stiffness, part of the pulse is reflected and the rest continues.
3. The probe records the time between emission and each returning echo.
4. A computer uses these times, and an assumed speed in tissue, to find the depth of each boundary and build an image.

Because the pulse goes there and back, depth = (speed × time delay) / 2.

Higher frequency means shorter wavelength, which lets you see finer detail. The cost is that higher frequencies are absorbed more strongly, so they do not reach as deeply. The choice of frequency is a compromise between detail and depth.

Worked example 5: depth of an organ

Assume ultrasound moves through soft tissue at 1540 m s⁻¹. Echoes from the front and back surfaces of an organ arrive 39 μs and 91 μs after the pulse is emitted. Find the thickness of the organ, and the wavelength at 3.5 MHz.

front: d₁ = 1540 × 39 × 10⁻⁶ / 2 = 0.0300 m
back:  d₂ = 1540 × 91 × 10⁻⁶ / 2 = 0.0701 m
thickness = 0.0701 - 0.0300 = 0.0400 m = 4.0 cm
λ = c / f = 1540 / 3.5 × 10⁶ = 4.4 × 10⁻⁴ m = 0.44 mm

Details much smaller than the wavelength cannot be resolved.

Polarisation

An unpolarised transverse wave oscillates in many directions perpendicular to the direction of travel. A plane-polarised wave oscillates in only one direction, so all the displacement (or electric field) lies in a single plane containing the direction of travel.

A Polaroid sheet has a transmission axis. It lets through the component of the oscillation along that axis and absorbs the component at right angles. Light passing through is plane-polarised along the axis.

Two Polaroid sheets in sequence show the effect:

• Unpolarised light through the first sheet becomes plane-polarised, with reduced intensity.
• Rotating the second sheet reduces the transmitted intensity.
• With the axes at 90° (crossed) almost no light passes.
• Rotating the second sheet through a full 360° gives two positions of maximum and two of minimum intensity.

Why this is evidence for transverse waves

Polarisation only makes sense if the oscillation has a direction at right angles to the direction of travel which can be selected or blocked. A longitudinal wave oscillates along the line of travel, so there is no sideways direction to filter, and rotating a filter would change nothing. Light, microwaves and radio waves can be polarised, which shows they are transverse. Sound cannot be polarised, which agrees with it being longitudinal.

Aerials

Radio and television signals are often transmitted with the electric field oscillating in one plane. A rod aerial responds to the component of the electric field along its length. The receiving aerial should therefore be parallel to the transmitting aerial:

• Vertical transmitter, vertical receiver: strongest signal.
• Vertical transmitter, horizontal receiver: almost no signal.
• Receiver tilted between the two: a reduced signal, from the component along the rod.

Turning a vertical rod about its own axis changes nothing, because the rod stays parallel to the oscillation.

Common errors

• Writing that particles “travel with the wave”. They oscillate about a fixed position.
• Defining transverse as “up and down”. Say at right angles to the direction the energy moves.
• Reading wavelength from a displacement-time graph, or period from a displacement-distance graph.
• Forgetting that ultrasound echo times include the journey there and back.
• Saying polarisation “proves” light is a wave in general. It is evidence that it is a transverse wave.
• Missing unit conversions: MHz, μs, nm.

Next steps

Test yourself with the practice questions, then use the revision notes to consolidate. The exam preparation guide covers technique, and the free diagnostics show which topics need work.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.5.3 and 3.5.4.