What are the main differences between transverse and longitudinal waves?

Transverse waves are characterized by particle displacement perpendicular to the direction of wave propagation, such as light waves. Longitudinal waves, like sound waves, have particle displacement parallel to the direction of wave propagation. Understanding these differences is crucial for Edexcel International A Levels Physics, especially in the Waves and Electricity section.

How do transverse waves and longitudinal waves propagate through different mediums?

Transverse waves require a medium that can support shear stress, such as solids, while longitudinal waves can travel through solids, liquids, and gases. This distinction is important in Edexcel International A Levels Physics, as it affects how waves are studied in different contexts.

Can you provide examples of transverse and longitudinal waves?

Common examples of transverse waves include electromagnetic waves like light and radio waves. Sound waves in air are a typical example of longitudinal waves. Recognizing these examples helps students in Edexcel International A Levels Physics to relate theoretical concepts to real-world phenomena.

Why is it important to understand wave properties in the context of Edexcel International A Levels Physics?

Understanding wave properties is fundamental to the Waves and Electricity section of Edexcel International A Levels Physics. It allows students to analyze how waves interact with different materials, which is essential for topics like optics, acoustics, and electromagnetic theory.

What role do transverse and longitudinal waves play in technological applications?

Transverse waves are crucial in technologies like fiber optics and wireless communication, while longitudinal waves are essential in sonar and medical ultrasound. These applications highlight the practical importance of wave studies in Edexcel International A Levels Physics, bridging theoretical knowledge with technological advancements.

Why is "Calling a sound wave a transverse wave (or describing sound as oscillating up and down)" a common mistake in Transverse and Longitudinal Waves?

Why it happens: Sound is almost always drawn as a sine curve, which looks transverse, so students copy that picture rather than thinking about how the air actually moves. How to avoid it: Remember that **sound is longitudinal**: the air particles oscillate back and forth ALONG the direction the wave travels, forming compressions and rarefactions. The sine curve only represents the pressure (or displacement) variation, not the actual particle motion. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Reading the wavelength off a displacement–time graph (or the period off a displacement–position graph)" a common mistake in Transverse and Longitudinal Waves?

Why it happens: The two graphs look identical (both sine curves), so students do not check whether the x-axis is time or distance. How to avoid it: **Always read the x-axis label first.** A displacement–TIME graph gives the **period** T; a displacement–POSITION (distance) graph gives the **wavelength** . Get f from T, then combine with in v=f. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Assuming the frequency of a wave changes when it passes from one medium into another" a common mistake in Transverse and Longitudinal Waves?

Why it happens: Students see the speed and wavelength change and assume the frequency must change too. How to avoid it: The **frequency is fixed by the source** and does NOT change between media. When the speed changes, it is the **wavelength** that changes to keep v=f consistent (=v/f). Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Quoting a phase difference as a bare fraction without converting to radians or degrees when asked" a common mistake in Transverse and Longitudinal Waves?

Why it happens: Students stop at 'a quarter of a cycle' and forget that one cycle is 2 rad or 360°. How to avoid it: Convert deliberately: =(x/)×2 rad (or ×360°). **One wavelength = 2 rad = 360°.** State the unit the question asks for (radians, degrees, or fraction of a cycle). Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Treating intensity as inversely proportional to distance (1/r) instead of to distance squared (1/r²)" a common mistake in Transverse and Longitudinal Waves?

Why it happens: Students remember that intensity 'gets weaker with distance' but forget the area of a sphere grows as r^2. How to avoid it: For a point source, I=P/(4 r^2), so **I\propto1/r^2**. Doubling the distance divides the intensity by 2^2=4; tripling it divides by 3^2=9. Use the sphere area 4 r^2, never r^2. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Forgetting that an echo (or sonar/CP4 reflection) involves the sound travelling there AND back — omitting the factor of two" a common mistake in Transverse and Longitudinal Waves?

Why it happens: Students apply v=d/t with d as the single distance to the reflector, ignoring the return trip. How to avoid it: For a reflected pulse the **total path is twice the distance to the reflector**: 2d=vt, so d=vt/2. Always sketch the out-and-back path before substituting. Source: Edexcel IAL Physics Unit 2 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Transverse and Longitudinal Waves

0% Complete

3 tasks remaining

Short Notes - Transverse and Longitudinal Waves

Page 1/0

Detailed Study Notes

Detailed notes on Waves and Electricity for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Transverse and Longitudinal Waves — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 2)

The language of waves (amplitude, wavelength, frequency, period, speed); the wave equation $v=f\lambda$ and $f=1/T$ ; transverse versus longitudinal waves with everyday examples; the crucial difference between a displacement–time and a displacement–position graph; wavefronts, phase and path difference; intensity $I=P/A$ and the inverse-square law; and Core Practical 4 — measuring the speed of sound in air.

At a glance

Transverse waves oscillate perpendicular to the energy transfer (all EM waves, string waves); longitudinal waves oscillate parallel to it (sound — compressions and rarefactions).
The wave equation: $v = f\lambda$ . Period and frequency are reciprocals: $f = \dfrac{1}{T}$ .
Amplitude = maximum displacement from equilibrium; wavelength = distance between adjacent in-phase points (crest to crest).
The graph trap: a displacement–time graph gives the period; a displacement–position graph gives the wavelength. Always read the x-axis label first.
Phase difference $\Delta\phi = \dfrac{x}{\lambda}\times 2\pi$ . One wavelength apart = in phase ( $2\pi$ ); half a wavelength apart = antiphase ( $\pi$ ).
Intensity $I = \dfrac{P}{A}$ (W m⁻²). For a point source $I = \dfrac{P}{4\pi r^2}$ , so $I\propto\dfrac{1}{r^2}$ — double the distance, quarter the intensity.
Frequency is set by the source and does NOT change between media; the wavelength and speed change so that $v=f\lambda$ still holds.
Core Practical 4: signal generator + speaker + microphone + double-beam oscilloscope; move the mic one wavelength to return the traces to being in phase, then $v=f\lambda$ .

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

2.1 — Use the terms amplitude, frequency, period, speed and wavelength to describe a wave, and recall that they are related by the wave equation $v=f\lambda$ and by $f=1/T$ .
2.2 — Distinguish between transverse and longitudinal waves in terms of the direction of oscillation relative to the direction of energy transfer, and give examples of each.
2.3 — Interpret displacement–time and displacement–position graphs of a wave, and recognise which can be read from each.
2.4 — Explain and use the concepts of wavefront, phase and phase difference, and path difference, including in-phase and antiphase relationships.
2.5 — Define intensity as power per unit area, $I=P/A$ , and apply the inverse-square relationship $I\propto1/r^2$ for a point source.
CP4 — Determine the speed of sound in air using a signal generator, speaker, microphone and oscilloscope, and use $v=f\lambda$ to obtain a value.

The language of waves and the wave equation

Five words and two equations describe every wave you will meet.

A progressive (travelling) wave transfers energy from one place to another without transferring matter — the particles of the medium simply oscillate about fixed positions. Five quantities describe any wave, and getting their definitions exact earns easy marks.

Amplitude ( $A$ ) — the maximum displacement of a particle from its equilibrium (rest) position. Larger amplitude ⇒ more energy.
Wavelength ( $\lambda$ ) — the distance between two adjacent points that are in phase (e.g. crest to crest, or compression to compression), in metres.
Frequency ( $f$ ) — the number of complete oscillations passing a point per second, in hertz (Hz).
Period ( $T$ ) — the time for one complete oscillation, in seconds.
Speed ( $v$ ) — the distance a wavefront travels per second, in m s⁻¹.

Two equations connect them. First, frequency and period are reciprocals: $f = \frac{1}{T}$

Second — the single most-used equation in this unit — the wave equation: $v = f\lambda$

This follows directly from "speed = distance ÷ time": in one period $T$ the wave advances exactly one wavelength $\lambda$ , so $v = \dfrac{\lambda}{T} = f\lambda$ .

For a transverse wave, the amplitude is the maximum displacement from the equilibrium position, and the wavelength is the distance between adjacent in-phase points such as two successive crests.

A useful consequence of $v=f\lambda$ : when a wave passes from one medium into another (e.g. sound from water into air, or light into glass), the frequency stays the same because it is fixed by the source — it is the speed and the wavelength that change together.

Amplitude = maximum displacement from equilibrium; wavelength = adjacent in-phase points.
$f = 1/T$ and the wave equation $v = f\lambda$ .
In one period the wave advances exactly one wavelength.
Frequency is set by the source and is unchanged between media; speed and wavelength change.

See the full worked example for transverse and longitudinal waves →

Transverse vs longitudinal waves

It comes down to one question: do the particles wobble across the wave, or along it?

Waves are classified by the direction the particles oscillate relative to the direction the energy travels.

Transverse wave — the oscillations are perpendicular (at $90°$ ) to the direction of energy transfer. The wave has visible crests and troughs. Examples: all electromagnetic waves (light, radio, microwaves, X-rays), waves on a stretched string, and water surface waves. A key property: transverse waves can be polarised.
Longitudinal wave — the oscillations are parallel to the direction of energy transfer. The medium forms regions where particles bunch together (compressions, high pressure) and regions where they spread apart (rarefactions, low pressure). The classic example is a sound wave in air (or any fluid). Primary (P) seismic waves are also longitudinal.

In a longitudinal wave (e.g. sound) the particles oscillate back and forth along the direction of energy transfer, creating compressions (high pressure, particles close) and rarefactions (low pressure, particles spread). One wavelength runs from one compression to the next.

Sound is often drawn as a smooth sine curve, which makes it look transverse — but that curve only represents how the pressure (or displacement) varies along the wave. The air itself oscillates back and forth along the direction of travel, which is why sound is longitudinal.

Transverse: oscillation perpendicular to energy transfer (EM waves, strings, water).
Longitudinal: oscillation parallel to energy transfer (sound — compressions/rarefactions).
Only transverse waves can be polarised.
Sound is longitudinal even though it is drawn as a sine curve.

See the full worked example for transverse and longitudinal waves →

The two graphs — and the trap examiners love

Same-looking sine curves, completely different x-axes: one gives the period, the other the wavelength.

Two graphs are used to represent a wave, and they look identical — both are sine curves. The difference is entirely in the horizontal axis, and reading the wrong one is the single most common error in this topic.

Displacement–time graph — the x-axis is time. It follows one particle as the wave passes. The horizontal repeat distance is therefore the period $T$ (the time for that particle to complete one oscillation). You cannot read the wavelength from it.
Displacement–position graph (also called displacement–distance) — the x-axis is position along the wave. It is a snapshot of the whole wave frozen at one instant. The horizontal repeat distance is therefore the wavelength $\lambda$ . You cannot read the period from it.

The two graphs are deliberately similar. The displacement–time graph (left) tracks one particle, so its repeat distance is the period T. The displacement–position graph (right) is a snapshot of the whole wave, so its repeat distance is the wavelength λ. Always check the x-axis label before reading anything off.

Putting them together. A typical exam question gives you both graphs and asks for the wave speed. The method is:

Read the period $T$ from the displacement–time graph, then $f = 1/T$ .
Read the wavelength $\lambda$ from the displacement–position graph.
Combine: $v = f\lambda$ .

Displacement–time graph: x-axis is time → gives the period $T$ .
Displacement–position graph: x-axis is distance → gives the wavelength $\lambda$ .
Get frequency from the period ( $f=1/T$ ), then $v=f\lambda$ .
Always read the x-axis label before measuring a repeat distance.

See the full worked example for transverse and longitudinal waves →

Wavefronts, phase and path difference

Phase tells you where in the cycle a particle is; phase difference compares two of them.

A wavefront is a line (or surface) joining all the points of a wave that are in phase — for example a line along all the crests. Wavefronts are perpendicular to the direction of energy transfer and are spaced one wavelength apart.

Phase describes the stage a particle has reached in its oscillation cycle. Because one complete cycle is $360°$ or $2\pi$ radians, phase is measured as a fraction of a cycle, in degrees, or in radians.

Phase difference ( $\Delta\phi$ ) compares two oscillating particles. It depends on their separation $x$ along the wave as a fraction of a wavelength: $\Delta\phi = \frac{x}{\lambda}\times 2\pi \quad (\text{in radians})$

Two special cases recur constantly:

In phase — separation is a whole number of wavelengths ( $x = 0, \lambda, 2\lambda,\ldots$ ), so $\Delta\phi = 0, 2\pi, 4\pi,\ldots$ . The particles move identically (both at a crest at the same time).
Antiphase (exactly out of phase) — separation is an odd number of half-wavelengths ( $x = \tfrac{\lambda}{2}, \tfrac{3\lambda}{2},\ldots$ ), so $\Delta\phi = \pi, 3\pi,\ldots$ . When one is at a crest, the other is at a trough.

Two waves a quarter of a wavelength apart have a path difference of λ/4 and a phase difference of π/2 rad (90°). A whole-wavelength gap would put them in phase (2π); a half-wavelength gap would put them in antiphase (π).

Path difference is the closely related idea used when two waves travel different distances to reach the same point: it is the difference in those distances, often quoted as a number of wavelengths. A path difference of one whole wavelength corresponds to a phase difference of $2\pi$ — this is exactly the principle behind Core Practical 4.

Wavefronts join in-phase points, are perpendicular to energy transfer, and are one wavelength apart.
$\Delta\phi = (x/\lambda)\times2\pi$ rad; one cycle $=360°=2\pi$ rad.
In phase: whole number of wavelengths apart ( $\Delta\phi=2\pi n$ ).
Antiphase: odd number of half-wavelengths apart ( $\Delta\phi=\pi, 3\pi,\ldots$ ).

See the full worked example for transverse and longitudinal waves →

Intensity and the inverse-square law

Power spread over area; for a point source it thins out as $1/r^2$ .

The intensity of a wave is the power transmitted per unit area at right angles to the direction of energy transfer: $I = \frac{P}{A}$ with $P$ in watts, $A$ in m², so intensity is measured in W m⁻². Intensity is also proportional to the square of the amplitude ( $I\propto A^2$ ), which is why doubling the amplitude of a wave quadruples its intensity.

Point sources and the inverse-square law. A small source radiating uniformly in all directions spreads its power over the surface of a sphere of radius $r$ , whose area is $A = 4\pi r^2$ . Substituting: $I = \frac{P}{4\pi r^2}\quad\Rightarrow\quad I\propto\frac{1}{r^2}$

So intensity falls off with the square of the distance. Key consequences to memorise:

Double the distance ( $r\to2r$ ) ⇒ intensity falls to $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$ .
Triple the distance ( $r\to3r$ ) ⇒ intensity falls to $\left(\tfrac{1}{3}\right)^2 = \tfrac{1}{9}$ .

This explains why a distant star or a far-off loudspeaker appears so faint: the same power is shared over an ever-larger spherical shell.

$I = P/A$ , measured in W m⁻²; $I\propto A^2$ (amplitude squared).
Point source: $A=4\pi r^2$ , so $I=P/(4\pi r^2)$ .
Inverse-square law: $I\propto1/r^2$ .
Double the distance → quarter the intensity; triple → one ninth.

See the full worked example for transverse and longitudinal waves →

Core Practical 4 — the speed of sound in air

Compare phases on an oscilloscope; one wavelength of travel realigns the traces.

Core Practical 4 determines the speed of sound in air using the wave equation $v=f\lambda$ . A reliable phase-comparison method uses four pieces of apparatus:

a signal generator set to a known frequency $f$ (typically a few kHz),
a loudspeaker driven by the signal generator,
a microphone that detects the sound,
a double-beam (dual-trace) oscilloscope showing both the source signal and the microphone signal at once.

Method. Connect the signal generator to the speaker and to one beam of the oscilloscope; connect the microphone to the other beam. Place the microphone next to the speaker and adjust its position until the two traces are exactly in phase (crests aligned). Now move the microphone slowly directly away from the speaker: the traces drift out of phase, through antiphase, and back into phase again. Between two successive in-phase positions the path has increased by exactly one wavelength, because the phase has changed by $2\pi$ . Measure that distance to get $\lambda$ .

Then apply the wave equation: $v = f\lambda$ with $f$ read from the signal generator and $\lambda$ from the microphone displacement. At room temperature this gives a value close to the accepted $\approx 340\ \text{m s}^{-1}$ .

Reducing uncertainty (worth ready marks):

Measure across many wavelengths — move the microphone through, say, ten in-phase positions and divide the total distance by ten. This reduces the percentage uncertainty in $\lambda$ far more than measuring a single wavelength.
Repeat and average, and work away from hard reflecting surfaces so reflected sound (standing waves) does not blur the in-phase positions.
Check $f$ on the oscilloscope's own time-base rather than trusting the dial alone, and ensure the microphone moves along the line directly away from the speaker.

Apparatus: signal generator + speaker + microphone + double-beam oscilloscope.
Move the mic one wavelength to return the two traces to being in phase.
Speed from $v=f\lambda$ ( $f$ from the generator, $\lambda$ measured).
Average over many wavelengths and avoid reflections to cut uncertainty.

See the full worked example for transverse and longitudinal waves →

Quick recap

A progressive wave transfers energy without transferring matter; describe it with amplitude, wavelength, frequency, period and speed.
Use $v=f\lambda$ and $f=1/T$ — convert prefixes (kHz, MHz, ms, cm) to SI units before substituting.
Transverse waves oscillate perpendicular to energy transfer (EM, strings); longitudinal waves oscillate parallel (sound — compressions and rarefactions).
Displacement–time graph → period; displacement–position graph → wavelength. Read the x-axis label first.
Phase difference $\Delta\phi=(x/\lambda)\times2\pi$ : in phase = whole number of wavelengths apart; antiphase = odd number of half-wavelengths apart.
Intensity $I=P/A$ (W m⁻²); for a point source $I=P/(4\pi r^2)$ so $I\propto1/r^2$ — double the distance, quarter the intensity.
Core Practical 4: move the microphone one wavelength to restore in-phase traces, then $v=f\lambda$ ; average over many wavelengths.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Wave equation: $v = f\lambda$ ; and $f = \dfrac{1}{T}$ .
Transverse: oscillation perpendicular to energy transfer (EM waves, strings). Longitudinal: oscillation parallel (sound — compressions and rarefactions).
Displacement–time graph gives the PERIOD; displacement–position graph gives the WAVELENGTH.
Phase difference $\Delta\phi = \dfrac{x}{\lambda}\times 2\pi$ ; one wavelength apart = in phase ( $2\pi$ ), half a wavelength apart = antiphase ( $\pi$ ).
Intensity $I = \dfrac{P}{A}$ (W m⁻²); point source $I = \dfrac{P}{4\pi r^2}$ , so $I\propto\dfrac{1}{r^2}$ and $I\propto A^2$ .
Frequency is set by the source and does not change between media — speed and wavelength change.
CP4: successive in-phase positions on the oscilloscope are one wavelength apart; then $v=f\lambda$ .

How it’s examined

Waves are core Unit 2 content and appear on the Unit 2 paper (WPH12, Waves and Electricity), which is structured with short calculations, definitions, graph interpretation and a practical/data question. Typical formats: define amplitude/wavelength/frequency or transverse/longitudinal (1–3 marks, AO1); single- and multi-step calculations using $v=f\lambda$ and $f=1/T$ , including rearrangement and prefix conversion (2–4 marks, M and A marks); a graph question where you must read the period from a displacement–time graph and the wavelength from a displacement–position graph and combine them (3–4 marks, AO3); phase- and path-difference problems quoted in radians, degrees or fractions of a cycle (2–3 marks); intensity and inverse-square-law calculations (2–5 marks); and a Core Practical 4 question on measuring the speed of sound (method, why one wavelength returns the traces to being in phase, and reducing uncertainty). ALWAYS show the equation, substitution with units and a 3 s.f. answer — ‘no working = risk zero’. For ‘show that’ work fully to the given value (one extra s.f.); for graphs check the x-axis label before reading anything off.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 2 examiner reports; Edexcel IAL Physics Unit 2 sample assessment materials and mark schemes. Last reviewed 2026-06-26.

Take this whole topic with you

Step-by-step worked examples — Transverse and Longitudinal Waves

Step-by-step solutions to past-paper-style questions on transverse and longitudinal waves, written exactly the way a tutor would explain them at the board.

Question type:

1Classifying transverse and longitudinal waves (2 marks)
Getting startedIdentify & classify• wave types, AO1, Unit 2
Question
State whether each of the following is a transverse or a longitudinal wave: (a) a sound wave in air, (b) a wave on a stretched string, (c) a microwave, (d) a primary (P) seismic wave through the Earth. (2 marks)
Step-by-step solution
1. Step 1
  A transverse wave oscillates perpendicular to the direction of energy transfer; a longitudinal wave oscillates parallel to it. Sound and P-waves are pressure (compression) waves, so they are longitudinal.
2. Step 2
  Waves on a string and all electromagnetic waves (including microwaves) oscillate perpendicular to the energy transfer, so they are transverse.
Answer
(a) longitudinal; (b) transverse; (c) transverse; (d) longitudinal.
Examiner tip
Mark scheme: 1 mark for any two correct, 2 marks for all four. The most common slip is calling a sound wave transverse — sound is a series of compressions and rarefactions, so the air oscillates back and forth along the direction the wave travels.
2Using the wave equation v = fλ (2 marks)
Getting startedDirect calculation• wave equation, AO2, Unit 2
Question
A water wave has a frequency of 2.5 Hz and a wavelength of 0.80 m. Calculate the speed of the wave. (2 marks)
Step-by-step solution
1. Step 1
  Select the wave equation, which links speed, frequency and wavelength.
  $v = f\lambda$
2. Step 2
  Substitute the values with units and evaluate.
  $v = 2.5 \times 0.80 = 2.0\ \text{m s}^{-1}$
Answer
$v = 2.0\ \text{m s}^{-1}$ .
Examiner tip
Mark scheme: (1) correct equation/substitution $v=2.5\times0.80$ (M); (2) $2.0\ \text{m s}^{-1}$ with unit (A). The frequency must be in hertz and the wavelength in metres before multiplying — mismatched units (e.g. cm) are the usual error.
3Linking period and frequency (2 marks)
Getting startedDirect calculation• frequency, period, AO2, Unit 2
Question
An oscilloscope shows that one complete cycle of a sound wave lasts 2.0 ms. Calculate the frequency of the wave. (2 marks)
Step-by-step solution
1. Step 1
  The frequency is the reciprocal of the period. First convert the period to seconds: $2.0\ \text{ms} = 2.0\times10^{-3}\ \text{s}$ .
  $f = \frac{1}{T}$
2. Step 2
  Substitute with the period in seconds and evaluate.
  $f = \frac{1}{2.0\times10^{-3}} = 500\ \text{Hz}$
Answer
$f = 500\ \text{Hz}$ .
Examiner tip
Mark scheme: (1) $f=1/T$ with $T$ in seconds (M); (2) $500\ \text{Hz}$ (A). Forgetting the millisecond-to-second conversion gives $f=0.5\ \text{Hz}$ , a factor of 1000 out — always convert prefixes before substituting.
4Rearranging the wave equation for wavelength (3 marks)
Building confidenceDirect calculation• wave equation, AO2, Unit 2
Question
A radio station broadcasts at a frequency of 95.8 MHz. Radio waves travel at the speed of light, $3.00\times10^{8}\ \text{m s}^{-1}$ . Calculate the wavelength of these radio waves. (3 marks)
Step-by-step solution
1. Step 1
  Start from the wave equation and rearrange to make wavelength the subject.
  $v = f\lambda \;\Rightarrow\; \lambda = \frac{v}{f}$
2. Step 2
  Convert the frequency to hertz: $95.8\ \text{MHz} = 95.8\times10^{6}\ \text{Hz}$ .
  $f = 95.8\times10^{6}\ \text{Hz}$
3. Step 3
  Substitute with units and evaluate to 3 s.f.
  $\lambda = \frac{3.00\times10^{8}}{95.8\times10^{6}} = 3.13\ \text{m}$
Answer
$\lambda = 3.13\ \text{m}$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) rearrange $\lambda=v/f$ ; (2) convert MHz to Hz (M); (3) $3.13\ \text{m}$ (A). Leaving the frequency as 95.8 (forgetting the $\times10^{6}$ ) gives a wildly wrong wavelength of $3.13\times10^{6}\ \text{m}$ — handle the powers of ten carefully.
5Reading period and wavelength from two graphs (4 marks)
Building confidenceGraph or diagram• graphs, wave equation, AO3, Unit 2
Question
For one travelling wave, a displacement–time graph for a single particle shows a period of 0.040 s, and a displacement–position (displacement–distance) graph at one instant shows a wavelength of 0.60 m. (a) State which graph gives the period and which gives the wavelength. (b) Calculate the speed of the wave. (4 marks)
Step-by-step solution
1. Step 1
  The displacement–time graph (x-axis is time) shows how one particle moves, so its repeat distance along the axis is the period $T$ . The displacement–position graph (x-axis is distance) is a snapshot of the whole wave, so its repeat distance is the wavelength $\lambda$ .
2. Step 2
  Find the frequency from the period read off the displacement–time graph.
  $f = \frac{1}{T} = \frac{1}{0.040} = 25\ \text{Hz}$
3. Step 3
  Use the wave equation with this frequency and the wavelength from the displacement–position graph.
  $v = f\lambda = 25 \times 0.60$
4. Step 4
  Evaluate the speed.
  $v = 15\ \text{m s}^{-1}$
Answer
(a) Period from the displacement–time graph; wavelength from the displacement–position graph. (b) $v = 15\ \text{m s}^{-1}$ .
Examiner tip
Mark scheme: (1) correct identification of each graph; (2) $f=25\ \text{Hz}$ ; (3) $v=f\lambda$ (M); (4) $15\ \text{m s}^{-1}$ (A). The classic error is reading the wavelength off a displacement–time graph (whose x-axis is time, not distance) — always check the axis label first.
6Phase difference between two points on a wave (3 marks)
Building confidenceMulti-step problem• phase difference, AO2, Unit 2
Question
A progressive wave has a wavelength of 1.2 m. Two points P and Q lie on the wave, separated by a distance of 0.30 m along the direction of travel. Calculate the phase difference between P and Q (a) as a fraction of a cycle and (b) in radians. (3 marks)
Step-by-step solution
1. Step 1
  Phase difference is the path difference expressed as a fraction of one wavelength. A whole wavelength corresponds to one full cycle.
  $\frac{\text{path difference}}{\lambda} = \frac{0.30}{1.2} = 0.25\ \text{of a cycle}$
2. Step 2
  One full cycle is $2\pi$ radians (or $360°$ ), so multiply the fraction of a cycle by $2\pi$ .
  $\Delta\phi = \frac{0.30}{1.2}\times 2\pi = 0.25 \times 2\pi$
3. Step 3
  Evaluate the phase difference in radians.
  $\Delta\phi = \frac{\pi}{2} = 1.57\ \text{rad}$
Answer
(a) $0.25$ of a cycle (a quarter of a cycle, equivalently $90°$ ). (b) $\Delta\phi = \tfrac{\pi}{2} = 1.57\ \text{rad}$ .
Examiner tip
Mark scheme: (1) fraction $=$ path difference $/\lambda = 0.25$ ; (2) multiply by $2\pi$ (M); (3) $1.57\ \text{rad}$ (A). Points exactly one wavelength apart are in phase ( $\Delta\phi=2\pi$ ); a half wavelength apart they are in antiphase ( $\Delta\phi=\pi$ ). A quarter wavelength gives $\pi/2$ .
7Intensity and the inverse-square law for a point source (5 marks)
StretchMulti-step problem• intensity, inverse square, AO2, Unit 2
Question
A small loudspeaker emits sound uniformly in all directions with a power of 0.50 W. (a) Calculate the intensity of the sound at a distance of 2.0 m from the speaker. (b) State, with a calculation, what the intensity becomes at 6.0 m. (5 marks)
Step-by-step solution
1. Step 1
  For a point source radiating uniformly, the power spreads over the surface of a sphere of radius $r$ , area $A=4\pi r^2$ . Intensity is power per unit area.
  $I = \frac{P}{A} = \frac{P}{4\pi r^2}$
2. Step 2
  Substitute for the 2.0 m distance.
  $I = \frac{0.50}{4\pi (2.0)^2} = \frac{0.50}{50.27}$
3. Step 3
  Evaluate the intensity at 2.0 m to 3 s.f.
  $I = 9.95\times10^{-3}\ \text{W m}^{-2}$
4. Step 4
  Intensity obeys an inverse-square law: $I\propto 1/r^2$ . Tripling the distance (2.0 m → 6.0 m) multiplies the intensity by $(1/3)^2=1/9$ .
  $I_{6} = \frac{I_{2}}{3^2} = \frac{9.95\times10^{-3}}{9}$
5. Step 5
  Evaluate the intensity at 6.0 m.
  $I_{6} = 1.11\times10^{-3}\ \text{W m}^{-2}$
Answer
(a) $I = 9.95\times10^{-3}\ \text{W m}^{-2}$ . (b) At 6.0 m, $I = 1.11\times10^{-3}\ \text{W m}^{-2}$ — one ninth of the value at 2.0 m.
Examiner tip
Mark scheme: (1) $I=P/(4\pi r^2)$ ; (2) substitution (M); (3) $9.95\times10^{-3}\ \text{W m}^{-2}$ (A); (4) inverse-square: factor $1/9$ (M); (5) $1.11\times10^{-3}\ \text{W m}^{-2}$ (A). Many candidates divide the intensity by 3 (the distance ratio) instead of by $3^2=9$ — intensity falls with the SQUARE of distance.
8Core Practical 4 — speed of sound from phase comparison (6 marks)
StretchMulti-step problem• Adapted from Edexcel IAL Physics Unit 2 (Core Practical 4)• Core Practical 4, speed of sound, AO3, Unit 2
Question
In Core Practical 4 a signal generator drives a loudspeaker at 2.0 kHz. A microphone is connected with the speaker to a double-beam oscilloscope and moved slowly away from the speaker. The two traces are first in phase; the next position at which they are again in phase is 0.171 m further from the speaker. (a) Explain why the microphone is moved one wavelength to return the traces to being in phase. (b) Calculate the speed of sound in air. (6 marks)
Step-by-step solution
1. Step 1
  The two oscilloscope traces are in phase whenever the microphone is a whole number of wavelengths from the speaker. Moving the microphone to the next in-phase position increases the path by exactly one wavelength, so the phase difference changes by $2\pi$ and the traces realign.
2. Step 2
  Therefore the distance moved equals one wavelength.
  $\lambda = 0.171\ \text{m}$
3. Step 3
  Convert the frequency to hertz: $2.0\ \text{kHz} = 2.0\times10^{3}\ \text{Hz} = 2000\ \text{Hz}$ .
  $f = 2000\ \text{Hz}$
4. Step 4
  Apply the wave equation with these values.
  $v = f\lambda = 2000 \times 0.171$
5. Step 5
  Evaluate the speed of sound to 3 s.f.
  $v = 342\ \text{m s}^{-1}$
Answer
(a) One wavelength of extra path changes the phase by $2\pi$ (one full cycle), bringing the traces back into phase. (b) $v = 342\ \text{m s}^{-1}$ — consistent with the accepted value of about $340\ \text{m s}^{-1}$ at room temperature.
Examiner tip
Mark scheme: (a) 2 marks — in phase when path difference is a whole number of wavelengths; next in-phase position is exactly one wavelength further. (b) $\lambda=0.171\ \text{m}$ ; convert kHz to Hz (M); $v=f\lambda$ (M); $342\ \text{m s}^{-1}$ (A). Measuring across several in-phase positions and dividing reduces the percentage uncertainty in $\lambda$ .
9Show that — distance to a cliff from an echo (6 marks)
StretchShow that / prove• speed of sound, echo, AO2, Unit 2
Question
A student claps once and hears a single echo from a cliff 0.96 s later. The speed of sound in air is 340 m s⁻¹. (a) Show that the cliff is about 160 m away. (b) Explain why the sound travels a total distance of twice the cliff separation. (6 marks)
Step-by-step solution
1. Step 1
  The sound travels to the cliff and back, so the total path length is $2d$ , where $d$ is the distance to the cliff. Use speed = distance ÷ time.
  $v = \frac{2d}{t} \;\Rightarrow\; 2d = v\,t$
2. Step 2
  Substitute the speed and the time with units.
  $2d = 340 \times 0.96 = 326.4\ \text{m}$
3. Step 3
  Halve the total path to get the distance to the cliff. Work to one extra significant figure for a 'show that'.
  $d = \frac{326.4}{2} = 163\ \text{m}$
4. Step 4
  This rounds to about 160 m, confirming the claim. The factor of two arises because the sound must travel out to the cliff and then reflect back to the listener.
Answer
(a) $d = 163\ \text{m} \approx 160\ \text{m}$ , as required. (b) The pulse travels to the cliff (distance $d$ ) and the reflected pulse returns the same distance $d$ , so the total path is $2d$ .
Examiner tip
Mark scheme: (a) recognise total path $=2d$ (M); $2d=v t=326\ \text{m}$ (M); $d=163\ \text{m}\approx160\ \text{m}$ (A) — quote one extra s.f. for 'show that'. (b) 2 marks: the sound goes to the cliff AND back, doubling the distance. Forgetting the factor of two (giving 326 m) is the most penalised error here.

Model Answers — Transverse and Longitudinal Waves

High-scoring sample answers for transverse and longitudinal waves on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define the amplitude and the wavelength of a progressive wave.
Model answer
The amplitude is the maximum displacement of a particle (or of the medium) from its equilibrium (rest) position.

The wavelength is the distance between two adjacent points on the wave that are in phase — for example from one crest to the next crest, or from one compression to the next compression.
Why this scores
Why this scores 2/2. (1) amplitude = maximum displacement from equilibrium; (2) wavelength = distance between adjacent in-phase points (crest-to-crest accepted). Defining wavelength loosely as 'the length of one wave' without 'in phase' or a clear crest-to-crest reference is not credited.
Question 2
3 marks
Q (3 marks). State one difference between a transverse and a longitudinal wave, and give one example of each.
Model answer
In a transverse wave the oscillations (vibrations) are perpendicular to the direction of energy transfer, whereas in a longitudinal wave the oscillations are parallel to the direction of energy transfer.

An example of a transverse wave is any electromagnetic wave (light, microwaves) or a wave on a string. An example of a longitudinal wave is a sound wave in air, which consists of compressions and rarefactions.
Why this scores
Why this scores 3/3. (1) transverse = oscillation perpendicular to energy transfer; (2) longitudinal = oscillation parallel to energy transfer; (3) one correct example of each. Stating 'energy transfer' (not 'the wave moves') is the precise wording examiners want.
Question 3
3 marks
Q (3 marks). A vibrating guitar string completes 250 oscillations in 0.50 s. Calculate (a) the frequency of the vibration and (b) the period of one oscillation.
Model answer
(a) Frequency is the number of complete oscillations per second:

$f = \frac{\text{number of oscillations}}{\text{time}} = \frac{250}{0.50} = 500\ \text{Hz}$

(b) The period is the reciprocal of the frequency:

$T = \frac{1}{f} = \frac{1}{500} = 2.0\times10^{-3}\ \text{s} = 2.0\ \text{ms}$
Why this scores
Why this scores 3/3. (a) 2 marks: $f=250/0.50=500\ \text{Hz}$ . (b) 1 mark: $T=1/f=2.0\times10^{-3}\ \text{s}$ . The period is the time for ONE oscillation, not the total 0.50 s — using the reciprocal of frequency avoids that confusion.
Question 4
4 marks
Q (4 marks). Sound travels through seawater at 1500 m s⁻¹. A sonar system on a ship emits a sound pulse of frequency 30 kHz. (a) Calculate the wavelength of this pulse in seawater. (b) The same pulse then travels into air, where the speed of sound is 340 m s⁻¹. State and explain what happens to its frequency and wavelength.
Model answer
(a) Rearranging the wave equation $v=f\lambda$ for wavelength, with $f=30\ \text{kHz}=3.0\times10^{4}\ \text{Hz}$ :

$\lambda = \frac{v}{f} = \frac{1500}{3.0\times10^{4}} = 0.050\ \text{m}$

(b) The frequency stays the same (30 kHz) because it is set by the source and does not change when a wave passes from one medium to another. Since the speed falls from 1500 m s⁻¹ to 340 m s⁻¹ while the frequency is fixed, $\lambda=v/f$ shows the wavelength decreases (to $340/3.0\times10^{4}=0.0113\ \text{m}$ ).
Why this scores
Why this scores 4/4. (a) 2 marks: $\lambda=v/f=0.050\ \text{m}$ (with kHz converted). (b) 2 marks: frequency unchanged (set by source) AND wavelength decreases because $v$ falls and $f$ is constant. A frequent error is assuming the frequency changes — it is the wavelength and speed that change between media.
Question 5
4 marks
Q (4 marks). A student is given two graphs of the same transverse water wave: graph A is a displacement–time graph for one floating cork, and graph B is a displacement–position graph showing the wave profile at one instant. (a) State which quantity can be read directly from each graph. (b) Graph A shows a period of 0.25 s and graph B shows a wavelength of 0.40 m. Calculate the speed of the wave.
Model answer
(a) Graph A (displacement against time) shows how one particle moves, so the period $T$ is read from the horizontal repeat distance — here $T=0.25\ \text{s}$ . Graph B (displacement against position/distance) is a snapshot of the whole wave, so the wavelength $\lambda$ is read from its horizontal repeat distance — here $\lambda=0.40\ \text{m}$ .

(b) First find the frequency from the period:

$f = \frac{1}{T} = \frac{1}{0.25} = 4.0\ \text{Hz}$

Then apply the wave equation:

$v = f\lambda = 4.0 \times 0.40 = 1.6\ \text{m s}^{-1}$
Why this scores
Why this scores 4/4. (a) 2 marks: period from the displacement–TIME graph, wavelength from the displacement–POSITION graph. (b) 2 marks: $f=1/T=4.0\ \text{Hz}$ then $v=f\lambda=1.6\ \text{m s}^{-1}$ . The discriminating skill is checking the x-axis label — reading 'wavelength' off a time-axis graph is the standard trap.
Question 6
5 marks
Q (5 marks). A progressive wave of wavelength 0.80 m travels along a rope. Points X and Y on the rope are separated by 0.20 m, and points X and Z are separated by 0.40 m. (a) Calculate the phase difference between X and Y in radians. (b) State the phase relationship between X and Z and justify your answer. (c) State the smallest separation, in terms of wavelength, for two points to be in phase.
Model answer
(a) The phase difference is the separation expressed as a fraction of a wavelength, multiplied by $2\pi$ :

$\Delta\phi = \frac{x}{\lambda}\times 2\pi = \frac{0.20}{0.80}\times 2\pi = 0.25 \times 2\pi = \frac{\pi}{2} = 1.57\ \text{rad}$

(b) X and Z are separated by 0.40 m, which is exactly half a wavelength ( $0.40/0.80=0.5$ ). Half a wavelength corresponds to a phase difference of $\pi$ rad ( $180°$ ), so X and Z are in antiphase (exactly out of phase): when X is at a crest, Z is at a trough.

(c) Two points are in phase when their separation is a whole number of wavelengths; the smallest non-zero separation is therefore one wavelength ( $1\lambda$ ).
Why this scores
Why this scores 5/5. (a) 2 marks: fraction $\times2\pi=1.57\ \text{rad}$ . (b) 2 marks: half a wavelength → $\pi$ rad → antiphase, with justification. (c) 1 mark: one whole wavelength. Stating numbers without the $2\pi$ conversion, or calling antiphase 'in phase', loses marks — antiphase means $\pi$ rad apart.
Question 7
5 marks
Q (5 marks). (a) Define the intensity of a wave and state its SI unit. (b) A lamp emits 12 W of light uniformly in all directions. Treating it as a point source, calculate the intensity of the light on a screen 3.0 m away.
Model answer
(a) The intensity of a wave is the radiant power transmitted per unit area at right angles to the direction of energy transfer:

$I = \frac{P}{A}$

Its SI unit is the watt per square metre, W m⁻².

(b) A point source radiates over the surface of a sphere of radius $r$ , area $A=4\pi r^2$ :

$I = \frac{P}{4\pi r^2} = \frac{12}{4\pi(3.0)^2} = \frac{12}{113.1} = 0.106\ \text{W m}^{-2}$
Why this scores
Why this scores 5/5. (a) 2 marks: power per unit area (defining equation $I=P/A$ ) plus unit W m⁻². (b) 3 marks: $A=4\pi r^2$ , substitution (M), $0.106\ \text{W m}^{-2}$ (A). Forgetting that the area is the surface of a SPHERE ( $4\pi r^2$ , not $\pi r^2$ ) is the most common slip.
Question 8
Adapted from Edexcel IAL Physics Unit 2 (Core Practical 4)6 marks
Q (6 marks). Describe how a signal generator, loudspeaker, microphone and a double-beam oscilloscope can be used to measure the speed of sound in air, and explain how the speed is calculated from the measurements. Include two ways of improving the accuracy of the result.
Model answer
Connect the signal generator to both the loudspeaker and one input of the double-beam oscilloscope, and connect the microphone to the second input. Set a known frequency $f$ (read from the signal generator) so the speaker emits a steady note. The oscilloscope shows two traces: the source signal and the microphone signal.

Move the microphone slowly directly away from the speaker. The two traces drift out of phase and then back into phase; the microphone has moved exactly one wavelength between successive in-phase positions, so measuring that distance gives the wavelength $\lambda$ .

The speed of sound is then found from the wave equation:

$v = f\lambda$

with $f$ from the signal generator and $\lambda$ from the microphone displacement.

Improving accuracy:

Measure across several wavelengths (e.g. the distance for ten successive in-phase positions) and divide by the number of wavelengths. This reduces the percentage uncertainty in $\lambda$ compared with measuring a single short wavelength.

Repeat the measurement and take a mean, and perform the experiment away from hard reflecting surfaces to avoid stray reflections (standing waves) that blur the in-phase positions.
Why this scores
Why this scores 6/6. Method (signal generator + speaker + mic + double-beam scope, in-phase comparison) and that one wavelength separates in-phase positions; $v=f\lambda$ with $f$ from the generator and $\lambda$ measured. Two genuine improvements WITH reasons — measuring many wavelengths reduces percentage uncertainty; repeats/avoiding reflections reduce error. Vague answers ('measure carefully') without the mechanism are not credited.
Question 9
Adapted from Edexcel IAL Physics Unit 26 marks
Q (6 marks). During a storm, a student sees a flash of lightning and hears the thunder 4.5 s later. The speed of sound in air is 340 m s⁻¹ and the speed of light is $3.0\times10^{8}\ \text{m s}^{-1}$ . (a) Estimate the distance to the lightning strike, justifying any assumption you make. (b) Explain, in terms of the type of wave, why the light arrives effectively instantly while the sound is delayed. (c) The thunder reaches the student at an intensity of $1.0\times10^{-4}\ \text{W m}^{-2}$ . State what would happen to this intensity if the student were twice as far from the strike, and justify your answer.
Model answer
(a) The light travels the distance almost instantly (it would take only about $10^{-5}\ \text{s}$ to cover a few kilometres), so the assumption is that the time delay is essentially the travel time of the sound alone. Then:

$d = v_{\text{sound}} \times t = 340 \times 4.5 = 1530\ \text{m} \approx 1.5\ \text{km}$

(b) Light is a transverse electromagnetic wave that travels at $3.0\times10^{8}\ \text{m s}^{-1}$ , almost a million times faster than sound. Sound is a longitudinal mechanical wave that needs a medium (air) and travels at only 340 m s⁻¹, so it arrives several seconds later — the gap between the flash and the thunder is the extra time the slower sound takes.

(c) Sound from the strike spreads out like a point source, so its intensity obeys an inverse-square law, $I\propto1/r^2$ . Doubling the distance multiplies the intensity by $(1/2)^2=1/4$ , so it would fall to:

$I = \frac{1.0\times10^{-4}}{4} = 2.5\times10^{-5}\ \text{W m}^{-2}$
Why this scores
Why this scores 6/6. (a) 2 marks: assume sound time dominates (light effectively instant) and $d=vt=1530\ \text{m}$ . (b) 2 marks: light = fast transverse EM wave, sound = slower longitudinal mechanical wave needing a medium. (c) 2 marks: inverse-square law, factor $1/4$ , $2.5\times10^{-5}\ \text{W m}^{-2}$ . Dividing the intensity by 2 instead of $2^2=4$ is the standard error.
Question 10
Adapted from Edexcel IAL Physics Unit 26 marks
Q (6 marks). A loudspeaker emits a single-frequency sound. A microphone records the wave, and the trace shows a period of 1.25 ms. The speed of sound in air is 340 m s⁻¹. (a) Calculate the frequency and the wavelength of the sound. (b) A second microphone is placed a distance further along the line from the speaker, where the sound is in antiphase with the first microphone. State the smallest such separation. (c) Explain why, when this sound passes from the air into a wall where its speed is greater, its wavelength increases but its frequency does not.
Model answer
(a) First convert the period: $T = 1.25\ \text{ms} = 1.25\times10^{-3}\ \text{s}$ . The frequency is:

$f = \frac{1}{T} = \frac{1}{1.25\times10^{-3}} = 800\ \text{Hz}$

Then the wavelength from the wave equation:

$\lambda = \frac{v}{f} = \frac{340}{800} = 0.425\ \text{m}$

(b) Two points are in antiphase when their separation is an odd number of half-wavelengths. The smallest such separation is half a wavelength:

$\frac{\lambda}{2} = \frac{0.425}{2} = 0.213\ \text{m}$

(c) The frequency is fixed by the source (the loudspeaker) and does not change when the wave enters a new medium — the number of oscillations arriving per second is unchanged. Since the speed increases in the wall while $f$ is constant, the wave equation rearranged as $\lambda = v/f$ shows that the wavelength must increase in proportion to the speed.
Why this scores
Why this scores 6/6. (a) 3 marks: convert ms to s, $f=800\ \text{Hz}$ , $\lambda=v/f=0.425\ \text{m}$ . (b) 1 mark: antiphase = half a wavelength = 0.213 m. (c) 2 marks: frequency set by source (unchanged); speed up with $f$ fixed → wavelength up via $\lambda=v/f$ . The synoptic discriminator is recognising that frequency, not wavelength, is conserved across a boundary.

Key Formulae — Transverse and Longitudinal Waves

The formulae you need to memorise for transverse and longitudinal waves on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

The wave equation
$v = f\lambda$
$v$
wave speed / m s⁻¹
$f$
frequency / Hz
$\lambda$
wavelength / m
When to use
Linking the speed, frequency and wavelength of any progressive wave. Rearrange to $\lambda=v/f$ or $f=v/\lambda$ as needed. In the Edexcel formulae booklet (Appendix 7).
Example
$f=2.5\ \text{Hz}$ , $\lambda=0.80\ \text{m}$ : $v=2.5\times0.80=2.0\ \text{m s}^{-1}$ .
Period–frequency relationship
$f = \frac{1}{T}$
$f$
frequency / Hz
$T$
period (time for one cycle) / s
When to use
Converting between the time for one complete oscillation (period) and the number of oscillations per second (frequency). Must be recalled.
Example
$T=2.0\ \text{ms}=2.0\times10^{-3}\ \text{s}$ : $f=1/(2.0\times10^{-3})=500\ \text{Hz}$ .
Intensity
$I = \frac{P}{A}$
$I$
intensity / W m⁻²
$P$
power (or radiant power) / W
$A$
area the power passes through / m²
When to use
Finding the power per unit area carried by a wave. For a point source radiating uniformly, $A=4\pi r^2$ , giving the inverse-square law $I\propto1/r^2$ . In the formulae booklet.
Example
$P=0.50\ \text{W}$ at $r=2.0\ \text{m}$ : $I=0.50/(4\pi(2.0)^2)=9.95\times10^{-3}\ \text{W m}^{-2}$ .
Phase difference from path difference
$\Delta\phi = \frac{x}{\lambda}\times 2\pi$
$\Delta\phi$
phase difference / rad
$x$
path difference (separation along the wave) / m
$\lambda$
wavelength / m
When to use
Converting a path difference (or separation of two points) into a phase difference. One whole wavelength corresponds to $2\pi$ rad (or $360°$ ). Must be recalled.
Example
$x=0.30\ \text{m}$ , $\lambda=1.2\ \text{m}$ : $\Delta\phi=(0.30/1.2)\times2\pi=\pi/2=1.57\ \text{rad}$ .
Wave speed from distance and time
$v = \frac{d}{t}$
$v$
wave speed / m s⁻¹
$d$
distance travelled by the wavefront / m
$t$
time taken / s
When to use
Timing-of-flight problems such as echoes, lightning–thunder gaps, or direct speed-of-sound measurements (remember total path = 2 × cliff distance for an echo). Must be recalled.
Example
Echo: $2d=v t=340\times0.96=326\ \text{m}$ , so $d=163\ \text{m}$ .

Key Definitions and Keywords — Transverse and Longitudinal Waves

Definitions to memorise and the exact keywords mark schemes credit for transverse and longitudinal waves answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Progressive (travelling) wave
Examiner keyword
A wave that transfers energy from one place to another through a medium (or through space) without any net transfer of matter. The particles oscillate about fixed equilibrium positions.
Transverse wave
Examiner keyword
A wave in which the oscillations (vibrations) are perpendicular to the direction of energy transfer. Examples: all electromagnetic waves, waves on a string, water surface waves.
Longitudinal wave
Examiner keyword
A wave in which the oscillations are parallel to the direction of energy transfer, producing compressions and rarefactions. Example: a sound wave in air.
Amplitude
Examiner keyword
The maximum displacement of a particle (or of the medium) from its equilibrium position. The greater the amplitude, the greater the energy and intensity of the wave.
Wavelength (λ)
Examiner keyword
The distance between two adjacent points on a wave that are in phase, e.g. crest to crest or compression to compression. Measured in metres.
Frequency (f)
Examiner keyword
The number of complete oscillations (or wavelengths) passing a fixed point per second. Measured in hertz (Hz); $f=1/T$ .
Period (T)
Examiner keyword
The time taken for one complete oscillation (or for one whole wavelength to pass a fixed point). Measured in seconds; $T=1/f$ .
Wave speed (v)
Examiner keyword
The distance travelled by a wavefront per unit time. Related to frequency and wavelength by the wave equation $v=f\lambda$ . Measured in m s⁻¹.
Compression
A region of a longitudinal wave where the particles are pushed close together, giving a higher-than-average pressure and density.
Rarefaction
A region of a longitudinal wave where the particles are spread apart, giving a lower-than-average pressure and density.
Wavefront
Examiner keyword
A line or surface joining all the points of a wave that are in phase (e.g. all the crests). Wavefronts are perpendicular to the direction of energy transfer and are spaced one wavelength apart.
Phase
Examiner keyword
The stage a particle has reached in its oscillation cycle, measured as a fraction of a cycle, in degrees, or in radians (one full cycle = $360°$ = $2\pi$ rad).
Phase difference
Examiner keyword
The difference in phase between two oscillating particles, given by $\Delta\phi=(x/\lambda)\times2\pi$ where $x$ is their separation. Quoted in radians, degrees, or fractions of a cycle.
In phase / antiphase
Examiner keyword
Two points are in phase if their phase difference is a whole number of cycles ( $0, 2\pi, 4\pi,\ldots$ ) — separated by a whole number of wavelengths. They are in antiphase if the phase difference is an odd number of half-cycles ( $\pi, 3\pi,\ldots$ ) — separated by an odd number of half-wavelengths.
Path difference
Examiner keyword
The difference in the distances travelled by two waves (or from one point to two others), measured in metres or as a number of wavelengths. A path difference of one wavelength corresponds to a phase difference of $2\pi$ .
Intensity (I)
Examiner keyword
The radiant power transmitted per unit area at right angles to the direction of energy transfer: $I=P/A$ , measured in W m⁻². For a point source, $I\propto1/r^2$ (inverse-square law).

Common Mistakes and Misconceptions — Transverse and Longitudinal Waves

The traps other students keep falling into on transverse and longitudinal waves questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Calling a sound wave a transverse wave (or describing sound as oscillating up and down)
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Sound is almost always drawn as a sine curve, which looks transverse, so students copy that picture rather than thinking about how the air actually moves.
How to avoid it
Remember that sound is longitudinal: the air particles oscillate back and forth ALONG the direction the wave travels, forming compressions and rarefactions. The sine curve only represents the pressure (or displacement) variation, not the actual particle motion.
✕Reading the wavelength off a displacement–time graph (or the period off a displacement–position graph)
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
The two graphs look identical (both sine curves), so students do not check whether the x-axis is time or distance.
How to avoid it
Always read the x-axis label first. A displacement–TIME graph gives the period $T$ ; a displacement–POSITION (distance) graph gives the wavelength $\lambda$ . Get $f$ from $T$ , then combine with $\lambda$ in $v=f\lambda$ .
✕Assuming the frequency of a wave changes when it passes from one medium into another
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students see the speed and wavelength change and assume the frequency must change too.
How to avoid it
The frequency is fixed by the source and does NOT change between media. When the speed changes, it is the wavelength that changes to keep $v=f\lambda$ consistent ( $\lambda=v/f$ ).
✕Quoting a phase difference as a bare fraction without converting to radians or degrees when asked
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students stop at 'a quarter of a cycle' and forget that one cycle is $2\pi$ rad or $360°$ .
How to avoid it
Convert deliberately: $\Delta\phi=(x/\lambda)\times2\pi$ rad (or $\times360°$ ). One wavelength = $2\pi$ rad = $360°$ . State the unit the question asks for (radians, degrees, or fraction of a cycle).
✕Treating intensity as inversely proportional to distance (1/r) instead of to distance squared (1/r²)
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students remember that intensity 'gets weaker with distance' but forget the area of a sphere grows as $r^2$ .
How to avoid it
For a point source, $I=P/(4\pi r^2)$ , so $I\propto1/r^2$ . Doubling the distance divides the intensity by $2^2=4$ ; tripling it divides by $3^2=9$ . Use the sphere area $4\pi r^2$ , never $\pi r^2$ .
✕Forgetting that an echo (or sonar/CP4 reflection) involves the sound travelling there AND back — omitting the factor of two
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students apply $v=d/t$ with $d$ as the single distance to the reflector, ignoring the return trip.
How to avoid it
For a reflected pulse the total path is twice the distance to the reflector: $2d=vt$ , so $d=vt/2$ . Always sketch the out-and-back path before substituting.

Transverse and Longitudinal Waves — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Transverse and Longitudinal Waves

Short Notes - Transverse and Longitudinal Waves

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Classifying transverse and longitudinal waves (2 marks)

2Using the wave equation v = fλ (2 marks)

3Linking period and frequency (2 marks)

4Rearranging the wave equation for wavelength (3 marks)

5Reading period and wavelength from two graphs (4 marks)

6Phase difference between two points on a wave (3 marks)

7Intensity and the inverse-square law for a point source (5 marks)

8Core Practical 4 — speed of sound from phase comparison (6 marks)

9Show that — distance to a cliff from an echo (6 marks)

The wave equation

Period–frequency relationship

Intensity

Phase difference from path difference

Wave speed from distance and time

Progressive (travelling) wave

Transverse wave

Longitudinal wave

Amplitude

Wavelength (λ)

Frequency (f)

Period (T)

Wave speed (v)

Compression

Rarefaction

Wavefront

Phase

Phase difference

In phase / antiphase

Path difference

Intensity (I)

✕Calling a sound wave a transverse wave (or describing sound as oscillating up and down)

✕Reading the wavelength off a displacement–time graph (or the period off a displacement–position graph)

✕Assuming the frequency of a wave changes when it passes from one medium into another

✕Quoting a phase difference as a bare fraction without converting to radians or degrees when asked

✕Treating intensity as inversely proportional to distance (1/r) instead of to distance squared (1/r²)

✕Forgetting that an echo (or sonar/CP4 reflection) involves the sound travelling there AND back — omitting the factor of two

Transverse and Longitudinal Waves — frequently asked questions