Waves — A-Level Physics Study Guide

For: A-Level Physics candidates sitting A-Level Physics.

Covers: Transverse vs longitudinal waves, core wave parameters, reflection/refraction/diffraction, polarisation of transverse waves, and the Doppler effect, fully aligned to the A-Level Physics syllabus.

You should already know: IGCSE Physics, basic algebra and trigonometry.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Physics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Waves?

A wave is a periodic disturbance that transfers energy from one point to another without transferring physical matter. Waves are ubiquitous in physics, ranging from sound waves that travel through air to electromagnetic waves (including light, radio waves, and X-rays) that can travel through a vacuum. This topic makes up 10-12% of total marks across A-Level Physics Papers 1, 2, and 4, with questions ranging from multiple-choice conceptual checks to multi-step calculation problems.

2. Transverse vs longitudinal waves

All mechanical waves fall into two core categories, defined by the relation between their oscillation direction and propagation direction:

• Transverse waves: The oscillation of particles (or fields, for electromagnetic waves) is perpendicular to the direction the wave travels. Examples include visible light, water surface waves, and waves travelling along a plucked guitar string. Transverse waves can be represented with peaks (maximum positive displacement) and troughs (maximum negative displacement) on a displacement-distance graph.
• Longitudinal waves: The oscillation of particles is parallel to the direction the wave travels. These waves form alternating regions of compression (particles packed closely together, high pressure) and rarefaction (particles spread apart, low pressure). Examples include sound waves, ultrasound, and seismic P-waves.

Worked Example

A student observes a wave moving along a slinky: individual coils move left and right horizontally as the wave travels right along the slinky. Classify the wave type. Solution: Oscillation direction is parallel to propagation direction, so the wave is longitudinal.

Exam tip: Examiners frequently ask for evidence that a wave is transverse: the only conclusive proof is that it can be polarised, as longitudinal waves cannot be polarised.

3. Wavelength, frequency, period, speed — $v=f\lambda$

Four core parameters define all waves, with a fixed relation between them:

1. Wavelength ($\lambda$): The distance between two consecutive identical points on a wave (e.g., peak to peak, compression to compression). SI unit: meters (m).
2. Period ($T$): The time taken for one full oscillation of a particle in the wave, or the time for one full wavelength to pass a fixed point. SI unit: seconds (s).
3. Frequency ($f$): The number of full oscillations per second, or the number of wavelengths passing a fixed point per second. SI unit: hertz (Hz, equivalent to $s^{-1}$). The relation between frequency and period is: $$f=\frac{1}{T}$$
4. Wave speed ($v$): The distance the wave travels per unit time. SI unit: $m{s}^{-1}$.

The wave speed formula is derived from the definition of speed: over one period $T$, the wave travels exactly one wavelength $\lambda$, so: $$v=\frac{\text{distance}}{\text{time}}=\frac{\lambda }{T}=f\lambda$$

Worked Example

A middle C note played on a piano has a frequency of 262 Hz, and travels through air at a speed of $330m{s}^{-1}$. Calculate its wavelength and period. Solution:

1. Wavelength: $\lambda =\frac{v}{f}=\frac{330}{262}\approx 1.26$ m
2. Period: $T=\frac{1}{f}=\frac{1}{262}\approx 3.82\times 10^{-3}$ s (or 3.8 ms)

Exam tip: Always convert all values to SI units before substituting into formulas. If a wavelength is given in cm, convert to m first: unit errors cost 1 mark per question even if your calculation logic is correct.

4. Reflection, refraction, diffraction

These three core wave behaviours apply to all wave types, and are frequently tested in diagram-based and conceptual questions:

• Reflection: When a wave hits an impenetrable boundary, it bounces back, following the law of reflection: the angle of incidence equals the angle of reflection, measured relative to the normal (a line perpendicular to the boundary). Examples include echoes (sound reflection) and mirror images (light reflection).
• Refraction: When a wave passes from one medium to another with a different density, its speed changes, causing a change in direction (unless the wave hits the boundary exactly along the normal). The frequency of the wave stays constant across both media (it is determined by the wave source), so wavelength changes proportionally to speed: $$\frac{v_1}{v_2}=\frac{{\lambda }_1}{{\lambda }_2}$$
• Diffraction: The spreading out of waves when they pass through a gap or travel around an obstacle. The extent of diffraction depends on the size of the gap relative to the wavelength of the wave: maximum diffraction occurs when the gap width is approximately equal to the wavelength. For example, you can hear sound around a closed door because sound wavelengths (~1 m) are similar to the width of a door gap, while visible light (wavelength ~500 nm) diffracts negligibly.

Worked Example

Light of wavelength 550 nm travels through air at $3\times 10^{8}m{s}^{-1}$, then enters a glass block with refractive index 1.5. Calculate its speed and wavelength in glass. Solution:

1. Frequency of light is constant: $f=\frac{c}{{\lambda }_{air}}=\frac{3\times 10^8}{550\times 10^{-9}}\approx 5.45\times 10^{14}$ Hz
2. Speed in glass: $v_{glass}=\frac{c}{n}=\frac{3\times 10^8}{1.5}=2\times 10^{8}m{s}^{-1}$
3. Wavelength in glass: ${\lambda }_{glass}=\frac{v_{glass}}{f}=\frac{2\times 10^8}{5.45\times 10^{14}}\approx 367$ nm

5. Polarisation of transverse waves

Polarisation is a phenomenon unique to transverse waves, where the oscillation of the wave is restricted to a single plane perpendicular to the direction of propagation. Longitudinal waves cannot be polarised, as their oscillation is parallel to the direction of travel, so there is no perpendicular plane to restrict.

• Unpolarised waves (e.g., natural sunlight) have oscillations in all random planes perpendicular to the direction of travel.
• A polariser is a filter that only allows waves oscillating along its polarisation axis to pass through. If unpolarised light of intensity $I_0$ passes through a polariser, the transmitted intensity is $\frac{I_0}{2}$, as only half the oscillation planes are allowed through.
• Malus's Law: If already polarised light of intensity $I_0$ passes through a second polariser (called an analyser) with its axis at angle $\theta$ to the incident polarisation direction, the transmitted intensity is: $$I=I_0{\cos }^2\theta$$

Worked Example

Polarised light of intensity $16W{m}^{-2}$ hits an analyser with its axis at 60° to the incident polarisation direction. Calculate the transmitted intensity. Solution: $I=16\times (\cos 60°{)}^2=16\times (0.5{)}^2=16\times 0.25=4W{m}^{-2}$

Real-world application: Polarising sunglasses block glare from roads and water, as reflected light is partially polarised horizontally. The sunglasses have a vertical polarisation axis, so they block all horizontally polarised reflected light.

6. Doppler effect — observed frequency change

The Doppler effect describes the change in observed frequency of a wave when the source or observer is moving relative to each other, even when the source emits a constant frequency. The effect occurs for all wave types, including sound and light.

• Intuition: If a source is moving towards an observer, wavefronts are compressed in front of the source, reducing wavelength and increasing observed frequency. If the source is moving away, wavefronts are stretched, increasing wavelength and reducing observed frequency. The same logic applies for a moving observer: an observer moving towards a source encounters more wavefronts per second, increasing observed frequency.
• For sound waves (the only quantitative version required for A-Level Physics), the observed frequency is given by: $$f_o=f_s\frac{v+v_o}{v-v_s}$$ Where $f_o$ = observed frequency, $f_s$ = source frequency, $v$ = speed of sound in the medium, $v_o$ = speed of the observer towards the source (negative if moving away), $v_s$ = speed of the source towards the observer (negative if moving away).
• For light, the Doppler effect causes red shift (lower observed frequency, longer wavelength) when a source is moving away from the observer, and blue shift (higher observed frequency, shorter wavelength) when moving towards the observer. Only a qualitative understanding is required for A-Level Physics.

Worked Example

An ambulance siren emits a sound of frequency 800 Hz, travelling towards a stationary pedestrian at 30 $m{s}^{-1}$. The speed of sound in air is 340 $m{s}^{-1}$. Calculate the frequency the pedestrian hears. Solution: The observer is stationary, so $v_o=0$. The source is moving towards the observer, so $v_s=+30m{s}^{-1}$. $f_o=800\times \frac{340+0}{340-30}=800\times \frac{340}{310}\approx 877$ Hz

Exam tip: Before substituting values, confirm if the observed frequency should be higher or lower than the source frequency to check your sign convention is correct.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Stating that longitudinal waves (e.g., sound) can be polarised. Why it happens: Confusing wave type properties. Fix: Memorise that only transverse waves can be polarised — this is the only definitive proof a wave is transverse.
• Pitfall 2: Using non-SI units in the $v=f\lambda$ formula, e.g., plugging wavelength in cm directly. Why it happens: Rushing through calculations. Fix: Convert all values to SI units (m, s, Hz, $m{s}^{-1}$) before substituting into any formula.
• Pitfall 3: Mixing up Doppler effect sign conventions, leading to lower observed frequency when the source is moving towards the observer. Why it happens: Memorising the formula without understanding the logic. Fix: First ask if $f_o$ should be higher or lower than $f_s$, then adjust signs to make the fraction larger/smaller accordingly.
• Pitfall 4: Stating that frequency changes during refraction. Why it happens: Confusing speed, wavelength, and frequency. Fix: Frequency is determined by the source, so it stays constant across media; only speed and wavelength change during refraction.
• Pitfall 5: Drawing diffracted wavefronts with a changed wavelength after a gap. Why it happens: Confusing diffraction with refraction. Fix: Wavelength stays identical after diffraction — only the shape of the wavefronts changes.

8. Practice Questions (A-Level Physics Style)

Question 1 (3 marks)

State two differences between transverse and longitudinal waves, and give one example of each type. Solution:

1. Difference 1: Transverse waves oscillate perpendicular to their direction of propagation, while longitudinal waves oscillate parallel to their direction of propagation (1 mark).
2. Difference 2: Transverse waves can be polarised, while longitudinal waves cannot (1 mark).
3. Examples: Transverse = light/electromagnetic waves; Longitudinal = sound waves (1 mark for both correct).

Question 2 (4 marks)

A FM radio station broadcasts at a frequency of 105 MHz. Electromagnetic waves travel at $3\times 10^{8}m{s}^{-1}$ in air. (a) Calculate the wavelength of the radio wave in air. (b) The wave enters a concrete wall where its speed falls to $2.1\times 10^{8}m{s}^{-1}$. State the frequency of the wave inside the wall, and calculate its new wavelength. Solution: (a) Step 1: Convert frequency to Hz: $105MHz=105\times 10^{6}Hz$ (1 mark). Step 2: $\lambda =\frac{v}{f}=\frac{3\times 10^8}{105\times 10^6}\approx 2.86m$ (1 mark). (b) Frequency remains constant at 105 MHz (1 mark). New wavelength: $\lambda =\frac{2.1\times 10^8}{105\times 10^6}=2.0m$ (1 mark).

Question 3 (5 marks)

A train horn emits sound of frequency 650 Hz, travelling away from a stationary railway worker at 40 $m{s}^{-1}$. The speed of sound in air is 330 $m{s}^{-1}$. (a) Calculate the frequency the worker hears. (b) Explain, in terms of wavefronts, why the observed frequency is different from the source frequency. Solution: (a) Step 1: Identify values: $v_o=0$, $v_s=-40m{s}^{-1}$ (source moving away from observer, so we use a negative sign for $v_s$, making the denominator $v+v_s$) (2 marks for correct sign convention). Step 2: $f_o=650\times \frac{330}{330+40}=650\times \frac{330}{370}\approx 580Hz$ (2 marks for correct answer with unit Hz). (b) As the train moves away from the worker, wavefronts emitted by the horn are stretched out behind the moving train, increasing the wavelength of the sound reaching the worker. For a fixed wave speed, increased wavelength leads to lower observed frequency (1 mark).

9. Quick Reference Cheatsheet

10. What's Next

Waves are a foundational topic that connects directly to multiple later units in the A-Level Physics syllabus: you will use the properties covered in this guide to understand superposition, interference, and stationary waves in Topic 8, electromagnetic spectrum behaviour in Topic 12, and quantum wave-particle duality in Topic 17. Mastering these concepts is critical to scoring well on the 10-15% of exam marks dedicated to wave-related questions across all papers, including practical assessment tasks on polarisation and diffraction.

If you struggle with any of the concepts, worked examples, or practice questions in this guide, you can get instant, personalised help from Ollie, our AI tutor, at any time by visiting Ollie. You can also access more A-Level Physics study guides, past paper walkthroughs, and custom practice quizzes on the homepage to build your confidence ahead of your exams.

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