Longitudinal and Transverse Waves

Understanding wave types, polarisation, and their applications

SPEC 3.3.1.2 • WAVES

Classifying Waves

In the previous lesson, we learned that waves transfer energy without transferring matter. Now we classify waves by the direction of oscillation relative to the direction of energy transfer.

Transverse Waves

Oscillations are perpendicular (at right angles) to the direction of energy transfer.

Longitudinal Waves

Oscillations are parallel to the direction of energy transfer.

Key Distinction

The critical difference is the direction of particle/field displacement relative to the direction the wave travels (energy propagation).

Transverse Waves

Definition

A transverse wave is a wave in which the oscillations are perpendicular (at right angles) to the direction of energy transfer.

Examples of transverse waves:

Wave TypeWhat OscillatesMedium

Electromagnetic wavesElectric and magnetic fieldsCan travel through vacuum

Waves on a stringParticles of stringSolid (string)

Water surface wavesWater moleculesLiquid surface

S-waves (seismic)Rock particlesSolid (Earth)

Longitudinal Waves

Definition

A longitudinal wave is a wave in which the oscillations are parallel to the direction of energy transfer.

Compression

Region where particles are close together (high density, high pressure).

Rarefaction

Region where particles are spread apart (low density, low pressure).

Examples: Sound waves, ultrasound, P-waves (seismic), pressure waves in fluids.

Worked Example01

During an earthquake, P-waves travel at 8 km/s and S-waves travel at 4.5 km/s. A seismometer detects P-waves 25 seconds before S-waves. Calculate the distance to the earthquake epicentre.

1

Let distance = d km. Time for P-waves: t_P = d/8. Time for S-waves: t_S = d/4.5

2

S-waves arrive 25s later: t_S - t_P = 25

3

d/4.5 - d/8 = 25

4

d(1/4.5 - 1/8) = 25 → d(0.222 - 0.125) = 25 → d(0.0972) = 25

Answerd = 25/0.0972 = 257 km (to 3 s.f.)

Worked Example02

A wave on a string travels horizontally from left to right. At one instant, a particle on the string is moving vertically upward. What type of wave is this? Justify your answer.

1

Direction of energy transfer: horizontal (left to right)

2

Direction of particle oscillation: vertical (upward at this instant)

3

The particle motion is perpendicular to the direction of energy transfer.

AnswerThis is a transverse wave because the particle oscillation (vertical) is perpendicular to the direction of energy transfer (horizontal).

Worked Example03

A student claims that water waves are purely transverse. Using your knowledge of wave types, evaluate this claim.

1

In deep water, particles move in circular paths - they have both vertical (up/down) and horizontal (back/forth) displacement.

2

The vertical component is perpendicular to wave travel → transverse behaviour.

3

The horizontal component is parallel to wave travel → longitudinal behaviour.

Conclusion: The claim is incorrect. Surface water waves are a combination of transverse and longitudinal motion - they are not purely transverse.

Transverse vs Longitudinal

PropertyTransverse WavesLongitudinal Waves

Oscillation directionPerpendicular to energy transferParallel to energy transfer

FeaturesCrests and troughsCompressions and rarefactions

Can be polarised?YesNo

Travel through vacuum?EM waves: YesNo (need medium)

ExamplesLight, waves on string, S-wavesSound, ultrasound, P-waves

Exam Tip

Polarisation is the key distinguishing feature. If a wave can be polarised, it must be transverse. Sound cannot be polarised, proving it is longitudinal.

Electromagnetic Waves

All electromagnetic (EM) waves are transverse waves. They consist of oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation.

Key Fact

All EM waves travel at the same speed in a vacuum: c = 3.0 × 10⁸ m/s

The Electromagnetic Spectrum

Remember

All EM waves: are transverse, travel at c = 3.0 × 10⁸ m/s in vacuum, can travel through a vacuum, and can be polarised. Since c is constant, higher frequency means shorter wavelength.

Worked Example04

Calculate the wavelength of a radio wave with frequency 98.5 MHz.

1

Convert frequency: f = 98.5 MHz = 98.5 × 10⁶ Hz

2

Use the wave equation: c = fλ

3

Rearrange: λ = c/f

4

Substitute: λ = (3.0 × 10⁸)/(98.5 × 10⁶) = 3.05 m

Answerλ = 3.05 m (or approximately 3.0 m)

Polarisation

Definition

Polarisation is the restriction of the oscillations of a transverse wave to a single plane.

Key Evidence

Only transverse waves can be polarised. Light can be polarised, proving it is transverse. Sound cannot be polarised, confirming it is longitudinal.

Worked Example05

Explain why sound waves cannot be polarised, while light waves can.

1

Polarisation is the restriction of oscillations to a single plane.

2

Sound waves are longitudinal - particles oscillate parallel to wave travel. There is only one possible direction of oscillation, so there is no plane to restrict.

3

Light waves are transverse - oscillations are perpendicular to travel direction. The electric field can oscillate in any plane perpendicular to travel, so a polariser can select one plane.

Key point: Only transverse waves can be polarised because they have multiple possible planes of oscillation.

Crossed Polarisers

When two polarising filters are at 90° to each other, no light passes through. This is called crossed polarisers.

Key Point

Only transverse waves can be polarised. If light were longitudinal, it would pass through crossed polarisers.

Worked Example06

Unpolarised light passes through two polarising filters at 90° to each other (crossed). Explain what is observed.

1

Unpolarised light has oscillations in all planes perpendicular to travel direction.

2

First polariser transmits only one plane - light becomes plane polarised (e.g., vertically).

3

Second polariser (analyser) at 90° can only transmit horizontally polarised light.

4

Vertically polarised light cannot pass through horizontal analyser.

Observation: No light transmitted - screen appears dark. This demonstrates light is a transverse wave.

Applications of Polarisers

Polaroid Sunglasses

Reduce glare from reflective surfaces. Light reflected from water/roads is partially horizontally polarised. Vertical polarising lenses block this.

LCD Screens

Use two polarising filters. Liquid crystals rotate polarisation. Controlling rotation controls pixel brightness.

3D Cinema

Two images with different polarisations. Glasses have different filters for each eye, creating depth perception.

Stress Analysis

Stressed materials rotate polarised light. Viewing through crossed polarisers reveals stress patterns.

Exam Focus

The specification mentions Polaroid material and aerial alignment as applications.

Aerial Alignment

Radio/TV signals are polarised EM waves. Transmitting and receiving aerials must be aligned in the same plane for maximum signal.

Why This Works

The oscillating E-field causes electrons in the aerial to oscillate. This only works when aerial is parallel to E-field direction. Perpendicular aerial receives no signal.

Worked Example07

Unpolarised light of intensity I₀ passes through a polarising filter. What is the intensity of the transmitted light? Explain your reasoning.

1

Unpolarised light has oscillations in all planes perpendicular to travel direction (random orientation).

2

A polariser only transmits the component of oscillation parallel to its transmission axis.

3

For light at angle θ to the axis, transmitted amplitude ∝ cos θ, so transmitted intensity ∝ cos²θ.

4

Averaging cos²θ over all angles from 0° to 360°: ⟨cos²θ⟩ = ½

AnswerTransmitted intensity = I₀/2 (half the original intensity)

Practice

Question 1: Seismic P-waves travel at 8 km/s, S-waves at 4.5 km/s. P-waves are longitudinal, S-waves are transverse. An earthquake occurs 200 km from a seismic station. Calculate the time difference between arrivals.

1

Time for P-waves: t_P = 200/8 = 25 s

2

Time for S-waves: t_S = 200/4.5 = 44.4 s

3

Time difference: Δt = 44.4 - 25 = 19.4 s

P-waves arrive first. This time difference helps locate earthquake epicentres.

Practice

Question 2: A microwave oven operates at 2.45 GHz. Calculate the wavelength. Explain why microwaves are transverse waves.

1

Convert: f = 2.45 GHz = 2.45 × 10⁹ Hz

2

λ = c/f = (3.0 × 10⁸)/(2.45 × 10⁹) = 0.122 m (≈ 12 cm)

3

Microwaves are EM waves. All EM waves have E and B fields perpendicular to propagation direction.

Perpendicular oscillation = transverse. They can also be polarised.

Practice

Question 3: Explain the physics behind how Polaroid sunglasses reduce glare from water or roads.

1

Light reflected from horizontal surfaces becomes partially horizontally polarised.

2

This horizontally polarised light is the "glare".

3

Polaroid sunglasses have vertically oriented polarising filters.

4

Vertical polarisers block horizontally polarised light (glare).

Result: Glare reduced while other light passes through.

Summary

ConceptKey Points

TransverseOscillations perpendicular; can be polarised; examples: EM waves, strings, S-waves

LongitudinalOscillations parallel; compressions/rarefactions; cannot be polarised; examples: sound, P-waves

EM wavesAll transverse; all travel at c = 3.0 × 10⁸ m/s in vacuum; can travel through vacuum

PolarisationRestriction to single plane; only transverse waves; evidence for light being transverse

ApplicationsPolaroid sunglasses, LCD screens, 3D cinema, aerial alignment

Exam Focus

Be able to: identify wave types, explain why only transverse waves can be polarised, describe applications, and explain aerial alignment.