What are the general properties of waves in Physics?

In Physics, particularly at the Cambridge IGCSE level, waves are characterized by several general properties. These include wavelength, frequency, amplitude, speed, and period. Wavelength is the distance between two consecutive points in phase on a wave, such as crest to crest. Frequency is the number of waves passing a point per second, measured in Hertz (Hz). Amplitude is the maximum displacement from the rest position, indicating the wave's energy. Speed is how fast the wave travels through a medium, and the period is the time taken for one complete wave cycle.

How do you differentiate between transverse and longitudinal waves?

Transverse and longitudinal waves differ in the direction of particle movement relative to wave propagation. In transverse waves, particles move perpendicular to the direction of wave travel, as seen in water waves or electromagnetic waves. In contrast, longitudinal waves have particles that move parallel to the wave direction, such as sound waves in air. Understanding these differences is crucial for Cambridge IGCSE Physics students studying wave properties.

What is the relationship between wave speed, frequency, and wavelength?

The relationship between wave speed (v), frequency (f), and wavelength (λ) is given by the equation v = f × λ. This equation shows that wave speed is the product of frequency and wavelength. For Cambridge IGCSE Physics students, this formula is essential for solving problems related to wave properties, allowing them to calculate any of the three variables if the other two are known.

Why is amplitude important in understanding wave properties?

Amplitude is a crucial property of waves because it indicates the wave's energy. A larger amplitude means more energy is carried by the wave. For example, in sound waves, a higher amplitude results in a louder sound. Understanding amplitude is vital for Cambridge IGCSE Physics students as it helps in analyzing the energy transfer and intensity of waves in various contexts.

How does the medium affect the speed of a wave?

The speed of a wave is significantly influenced by the medium through which it travels. Different media have varying densities and elastic properties, affecting wave speed. For instance, sound waves travel faster in solids than in liquids or gases due to the closer particle arrangement in solids. Cambridge IGCSE Physics students learn that the medium's properties, such as density and elasticity, determine how quickly waves can propagate through it.

Why is "Saying sound is a transverse wave" a common mistake in General Properties of Waves?

Why it happens: Default to transverse mental image. How to avoid it: Sound = LONGITUDINAL (compressions/rarefactions parallel to travel). Source: 0625/42 — recurring

Why is "Using kHz or MHz directly without converting" a common mistake in General Properties of Waves?

Why it happens: Frequencies in real questions use prefixes. How to avoid it: Convert to base Hz: 1\,kHz = 10^3\,Hz.

Why is "Saying gap size has no effect on diffraction" a common mistake in General Properties of Waves?

Why it happens: Confusing two diagrams in textbooks. How to avoid it: Diffraction is greatest when gap ≈ wavelength.

Why is "Confusing oscillation distance with wavelength" a common mistake in General Properties of Waves?

Why it happens: Misreading a snapshot diagram. How to avoid it: is the spatial period (crest to crest), not the displacement of a particle.

Cambridge IGCSE Physics (0625)

General Properties of Waves

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Short Notes - General Properties of Waves

Detailed Study Notes

Detailed notes on Waves for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

General Properties of Waves — Cambridge IGCSE 0625 Physics Extended (2026)

Transverse vs longitudinal, wavelength, frequency, period, $v = f\lambda$ , plus reflection, refraction and diffraction. The shared toolkit for sound, light and the EM spectrum.

At a glance

Transverse: oscillations PERPENDICULAR to direction of travel — EM radiation, water waves, seismic S-waves.
Longitudinal: oscillations PARALLEL to direction — sound waves and seismic P-waves.
Wavefront: a line joining points of a wave that are in step; adjacent wavefronts are one wavelength apart.
Wavelength $\lambda$ : distance for one full cycle ( $\text{m}$ ).
Frequency $f$ : cycles per second ( $\text{Hz}$ ). Period $T = 1/f$ .
Wave equation: $v = f\lambda$ .
Reflection / refraction / diffraction: angles equal; speed change at a boundary; spreading round gaps.

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

3.1 — Distinguish between transverse and longitudinal waves, using named examples (EM radiation, water and seismic S-waves as transverse; sound and seismic P-waves as longitudinal).
3.1 — Describe the features of a wave in terms of wavefront, wavelength, frequency, crest, trough and amplitude.
3.1 — Recall and use $v = f\lambda$ .
3.2 — Describe reflection, refraction and diffraction of waves.

Transverse and longitudinal waves

Transverse: peaks/troughs perpendicular to motion. Longitudinal: compressions/rarefactions along motion.

Transverse wave. The oscillation is PERPENDICULAR to the direction the wave travels. The named examples in the syllabus are electromagnetic radiation (light, radio, etc.), water waves and seismic S-waves (secondary waves) — all modelled as transverse.

Longitudinal wave. The oscillation is PARALLEL to the direction of travel — particles squeeze together (compressions) and stretch apart (rarefactions). The named examples are sound waves and seismic P-waves (primary waves) — both modelled as longitudinal.

Seismic waves are produced by earthquakes and travel through the Earth. Remember: P for Parallel vibration (longitudinal) and S for Sideways vibration (transverse).

Wavefronts. A wavefront is a line (or surface) joining all the points of a wave that are vibrating exactly in step — for example, a line joining a row of crests. Adjacent wavefronts are one wavelength apart. Drawing waves as a set of parallel wavefronts is a convenient way to show reflection, refraction and diffraction.

Both types share the features wavefront, wavelength, frequency, crest (peak), trough, amplitude and wave speed.

Drawing.

Transverse: a sine curve with crests (peaks) and troughs.
Longitudinal: a series of compressions (close particles) and rarefactions (spread particles).

Energy not matter. Waves transfer ENERGY without transferring matter. A floating cork bobs up and down as water waves pass — it doesn't travel with the wave.

Transverse waves oscillate across the direction of travel; longitudinal waves squeeze and stretch along it.

Transverse: oscillation $\perp$ travel — EM radiation, water, seismic S-waves.
Longitudinal: oscillation $\parallel$ travel — sound, seismic P-waves.
Wavefront: a line of in-step points; adjacent wavefronts one $\lambda$ apart.
Both transfer energy, not matter.

Wave parameters and the wave equation

$v = f\lambda$ . Memorise definitions of $\lambda, f, T$ .

Wavelength $\lambda$ : distance from one peak to the next (or one compression to the next). Units: $\text{m}$ .

Frequency $f$ : number of complete waves per second. Units: hertz ( $\text{Hz}$ ).

Period $T$ : time for one complete oscillation. $T = 1/f$ . Units: $\text{s}$ .

Amplitude $A$ : maximum displacement from rest position. Determines how much energy the wave carries.

Wave speed $v$ : how fast the wave travels. $\text{m/s}$ .

Wavelength λ is one full cycle (crest to crest); amplitude A is the maximum displacement from the rest position.

Wave equation. $v = f\lambda.$

Worked. A water wave has wavelength $0.5\,\text{m}$ and frequency $2\,\text{Hz}$ . Find speed.

$v = 2 \times 0.5 = 1\,\text{m/s}$ .

Worked. Sound in air at $340\,\text{m/s}$ , frequency $500\,\text{Hz}$ . Find wavelength.

$\lambda = v/f = 340/500 = 0.68\,\text{m}$ .

Tip. Always rearrange the equation to isolate the unknown — don't try to memorise three variants.

$v = f\lambda$ .
$T = 1/f$ .
Amplitude $\to$ energy carried.
Higher frequency $\to$ shorter wavelength (at fixed $v$ ).

Reflection and refraction

Reflection: angles equal. Refraction: speed change at boundary, direction bends.

Reflection. When a wave bounces off a surface.

Law of reflection: angle of incidence = angle of reflection (both measured from the NORMAL, not the surface).

Refraction. When a wave crosses into a different medium, its speed changes — and if it hits at an angle, its direction bends.

Going into a denser/slower medium → wave SLOWS down → bends TOWARD the normal.
Going into a less dense/faster medium → SPEEDS up → bends AWAY from normal.
Wavelength changes; frequency stays the same.

Worked qualitative. Light entering glass slows down → bends toward normal. Exiting back into air → speeds up → bends away from normal.

Tip. Always draw the normal (perpendicular) line at the boundary first. Both angles are measured from it.

Both angles are measured from the normal — never from the mirror surface.

Entering the slower medium the ray slows and bends toward the normal, so $r < i$; frequency is unchanged but wavelength shrinks.

Reflection: angles equal, measured from normal.
Refraction: speed AND direction change at boundary.
Slower medium: bend TOWARD normal.
Frequency stays constant; wavelength shrinks in slower medium.

Diffraction

Waves spread out as they pass through a gap or round an obstacle.

Diffraction. Waves bend or spread out when they pass through a gap or round an obstacle.

Amount of diffraction depends on the size of the gap relative to the wavelength:

Gap MUCH larger than $\lambda$ : little diffraction.
Gap $\approx \lambda$ : maximum diffraction.
Gap smaller than $\lambda$ : still spreads, but less wave gets through.

Why we hear round corners. Sound has long wavelengths (cm-m), so it diffracts well around doorways, walls, etc. Light has very short wavelengths (nm), so it doesn't diffract noticeably round corners — that's why you can't see round them.

Worked qualitative. A ripple tank with parallel plane waves hits a barrier with a small gap — circular waves spread out from the gap on the other side. With a wide gap, the waves continue mostly straight with only slight curving at the edges.

When the gap is about one wavelength wide, straight wavefronts emerge as circular ones — maximum diffraction.

Diffraction: waves spread round / through gaps.
Maximum when gap $\approx \lambda$ .
Sound diffracts well; light barely.
Demonstrated in ripple tanks.

Quick recap

Transverse (EM radiation, water, seismic S-waves) vs longitudinal (sound, seismic P-waves).
Wavefront = line of in-step points; adjacent wavefronts one $\lambda$ apart.
$v = f\lambda$ . $T = 1/f$ .
Reflection: angles equal (from normal).
Refraction: speed change → direction bend.
Diffraction: spreading; max when gap $\approx \lambda$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Transverse wave — oscillation perpendicular to direction of travel; e.g. EM radiation, water waves, seismic S-waves.
Longitudinal wave — oscillation parallel to direction of travel; e.g. sound waves, seismic P-waves.
Wavefront — a line joining points of a wave that are vibrating in step.
Wavelength — distance between two consecutive identical points on a wave.
Frequency — number of complete waves per second, $\text{Hz}$ .
Wave equation — $v = f\lambda$ .
Diffraction — spreading of waves around obstacles or through gaps.

How it’s examined

General wave properties appear every Paper 2 (3-5 marks: identify wave type, $v = f\lambda$ calculation) and Paper 4 (multi-part on ripple tank diffraction or refraction at a boundary). Examiner reports flag measuring angles from the surface and forgetting that frequency does NOT change at a boundary.

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (3.1-3.2); 0625/42 Oct/Nov 2024 — Q13 (ripple tank diffraction); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

Step-by-step worked examples — General Properties of Waves

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

Question type:

Question patterns to master — General Properties of Waves

Almost every general properties of waves exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

Two of $v$ , $f$ , $\lambda$ (or a period) are given and the third is wanted — find the speed / wavelength / frequency. One formula, one answer.

How to approach it

Use $v=f\lambda$ (or $f=1/T$ ), rearrange for the unknown, convert all quantities to SI base units, then substitute.

Common trap

Examiner reports flag unconverted prefixes — using kHz or MHz directly, or leaving $\lambda$ in mm. Convert to Hz and m before substituting, and always state the unit.

Identify & classify

Recognise it by

A list of waves (water, sound, light) with an instruction to classify them as transverse or longitudinal.

How to approach it

Compare the direction of oscillation with the direction of energy travel: perpendicular → transverse; parallel (compressions and rarefactions) → longitudinal.

Common trap

Examiner reports flag sound being called transverse. Sound is longitudinal — it travels as compressions and rarefactions along the direction of travel.

Graph or diagram

Recognise it by

A displacement–distance or displacement–time graph, a ripple-tank wavefront diagram, or an instruction to sketch diffraction or reflection.

How to approach it

For displacement–distance read amplitude off the y-axis and wavelength as one full cycle; for displacement–time read the period as one cycle. For wavefront diagrams measure all angles from the normal.

Common trap

Examiner reports flag angles measured from the barrier instead of the normal, and confusing a particle's displacement with the wavelength on a snapshot graph.

Multi-step problem

Recognise it by

A wave crossing a boundary (refraction) where you must find a speed in one medium and a wavelength in another — two linked steps.

How to approach it

Apply $v=f\lambda$ in the first medium, carry the unchanged frequency across the boundary, then apply $v=f\lambda$ again in the second medium.

Common trap

Examiner reports flag candidates changing the frequency at a boundary. Frequency is fixed by the source — only $v$ and $\lambda$ change on refraction.

Show that / prove

Recognise it by

The stem says explain why or state which … and why — typically comparing how strongly two wavelengths diffract around an obstacle.

How to approach it

Argue from the rule that diffraction is greatest when the obstacle or gap is comparable in size to the wavelength, then apply it to each case and conclude.

Common trap

Claiming gap or obstacle size has no effect on diffraction — the amount of spreading depends strongly on the wavelength-to-gap ratio.

1Use the wave equation
CoreDirect calculation• v=fλ
Question
A water wave has frequency $4\,\text{Hz}$ and wavelength $0.5\,\text{m}$ . Find its speed.
Step-by-step solution
1. Step 1
  $v = f\lambda$ .
  $v = 4 \times 0.5 = 2\,\text{m/s}$
Answer
$2\,\text{m/s}$
2Period and frequency
CoreDirect calculation• period
Question
A wave has period $0.02\,\text{s}$ . Find its frequency.
Step-by-step solution
1. Step 1
  $f = 1/T$ .
  $f = \dfrac{1}{0.02} = 50\,\text{Hz}$
Answer
$50\,\text{Hz}$
3Distinguish transverse and longitudinal
CoreIdentify & classify• wave types
Question
Classify: water waves on a pond, sound in air, EM waves.
Step-by-step solution
1. Step 1
  Transverse: oscillation perpendicular to wave direction.
2. Step 2
  Longitudinal: oscillation parallel to wave direction (compressions and rarefactions).
Answer
Water and EM: transverse. Sound: longitudinal.
4Seismic P-waves and S-waves — transverse or longitudinal?
Core• wave types, seismic
Question
An earthquake produces two types of seismic wave: P-waves (primary) and S-waves (secondary). (a) State which of these can be modelled as transverse and which as longitudinal. (b) For each, describe the direction of vibration relative to the direction the wave travels.
Step-by-step solution
1. Step 1
  Recall the named examples: sound and seismic P-waves are modelled as longitudinal; electromagnetic radiation, water waves and seismic S-waves are modelled as transverse.
2. Step 2
  (a) Seismic P-waves are longitudinal; seismic S-waves are transverse.
3. Step 3
  (b) In a P-wave the rock particles vibrate back and forth PARALLEL to the direction the wave travels (compressions and rarefactions). In an S-wave the rock particles vibrate at RIGHT ANGLES (perpendicular) to the direction the wave travels.
Answer
(a) P-waves: longitudinal. S-waves: transverse. (b) P-wave vibrations are parallel to the direction of travel; S-wave vibrations are perpendicular to it.
Examiner tip
P-waves and S-waves are the named examples in the 2026 syllabus §3.1 — remember 'P for Parallel (longitudinal)' and 'S for Sideways (transverse)'.
5Describing a wavefront
Core• wavefront, wave types
Question
A ripple tank produces straight (plane) water waves. (a) Explain what is meant by a wavefront. (b) State how the distance between two adjacent wavefronts is related to the wavelength.
Step-by-step solution
1. Step 1
  (a) A wavefront is a line (or surface) joining all the points on a wave that are vibrating exactly in step — for example, a line joining all the crests of one ripple.
2. Step 2
  (b) Adjacent wavefronts are one wavelength apart, because successive crests are separated by one wavelength.
Answer
(a) A wavefront joins all points of a wave that are in phase (e.g. a line of crests). (b) The distance between two adjacent wavefronts equals one wavelength.
6Diffraction through a gap
ExtendedGraph or diagram• Adapted from 0625/42 May/Jun 2024 Q12• diffraction
Question
Sketch how diffraction changes when the gap is similar to the wavelength versus much larger than the wavelength.
Step-by-step solution
1. Step 1
  Gap ≈ wavelength → significant spreading (large diffraction).
2. Step 2
  Gap ≫ wavelength → almost no spreading (waves continue almost straight).
Answer
Maximum diffraction occurs when the gap width is similar to the wavelength.
7Find the wavelength from speed and frequency
ExtendedDirect calculation• Adapted from 0625/42 Oct/Nov 2023 Q13• v=fλ
Question
A sound wave in air travels at $340\,\text{m/s}$ with a frequency of $1700\,\text{Hz}$ . Find the wavelength.
Step-by-step solution
1. Step 1
  $v = f\lambda \Rightarrow \lambda = v/f$ .
  $\lambda = \dfrac{340}{1700} = 0.20\,\text{m}$
Answer
$\lambda = 0.20\,\text{m} = 20\,\text{cm}$
Examiner tip
The examiner report flags candidates often quote the answer in metres without unit conversion checks (e.g. mm vs m). Always state the unit explicitly.
8Read amplitude and wavelength from a graph
ExtendedGraph or diagram• graph, transverse
Question
A transverse wave on a string is drawn on a displacement-distance graph. The y-axis ranges $-3\,\text{cm}$ to $+3\,\text{cm}$ ; the wave shows two complete cycles between $x = 0\,\text{m}$ and $x = 1.0\,\text{m}$ . State the amplitude and wavelength.
Step-by-step solution
1. Step 1
  Amplitude = maximum displacement from rest.
  $A = 3\,\text{cm} = 0.03\,\text{m}$
2. Step 2
  Two cycles in $1.0\,\text{m}$ → one cycle in $0.50\,\text{m}$ .
  $\lambda = 0.50\,\text{m}$
Answer
$A = 0.03\,\text{m}$ ; $\lambda = 0.50\,\text{m}$
9Read period and frequency from a graph
ExtendedGraph or diagram• graph, period
Question
A displacement-time graph for a wave shows three complete oscillations between $t = 0$ and $t = 0.12\,\text{s}$ . Find the period and frequency.
Step-by-step solution
1. Step 1
  Period = time for one complete oscillation.
  $T = \dfrac{0.12}{3} = 0.04\,\text{s}$
2. Step 2
  $f = 1/T$ .
  $f = \dfrac{1}{0.04} = 25\,\text{Hz}$
Answer
$T = 0.04\,\text{s}$ ; $f = 25\,\text{Hz}$
10Reflection of plane waves at a barrier
ExtendedGraph or diagram• reflection, ripple
Question
Plane water waves in a ripple tank approach a straight barrier at $30\degree$ to the normal. State (a) the angle of reflection, (b) what happens to the wavelength, frequency and speed of the reflected wave.
Step-by-step solution
1. Step 1
  Angle of reflection = angle of incidence, both measured from the normal.
  $\theta_{r} = 30\degree$
2. Step 2
  Reflection does NOT change wavelength, frequency or speed — only direction.
Answer
(a) $30\degree$ from the normal. (b) Wavelength, frequency and speed are all unchanged.
Examiner tip
The examiner report flags candidates often measure the angle from the barrier itself. Always measure from the normal — the perpendicular to the barrier.
11Refraction of a wave at a boundary
ChallengeMulti-step problem• Adapted from 0625/42 May/Jun 2024 Q12• refraction, wave
Question
A water wave of frequency $5.0\,\text{Hz}$ has wavelength $40\,\text{mm}$ in deep water. When it crosses into shallow water its speed falls to $0.10\,\text{m/s}$ . Find (a) the speed in deep water, (b) the wavelength in shallow water.
Step-by-step solution
1. Step 1
  Speed in deep water.
  $v_{1} = f\lambda_{1} = 5.0 \times 0.040 = 0.20\,\text{m/s}$
2. Step 2
  Frequency is unchanged on crossing a boundary; only speed and wavelength change.
  $\lambda_{2} = \dfrac{v_{2}}{f} = \dfrac{0.10}{5.0}$
3. Step 3
  Compute.
  $\lambda_{2} = 0.020\,\text{m} = 20\,\text{mm}$
Answer
(a) $v_{1} = 0.20\,\text{m/s}$ . (b) $\lambda_{2} = 20\,\text{mm}$ .
Examiner tip
The examiner report flags candidates often change frequency at the boundary. Frequency is fixed by the source — only $v$ and $\lambda$ change on refraction.
12A* — Compare diffraction of long and short wavelengths
ChallengeShow that / prove• diffraction, synoptic
Question
A long-wave radio station broadcasts at $\lambda \approx 1500\,\text{m}$ and a TV transmitter broadcasts at $\lambda \approx 0.5\,\text{m}$ . Both encounter a hill $30\,\text{m}$ wide blocking the line of sight to a house behind it. (a) State which signal is received more strongly behind the hill and why. (b) Use this to suggest one practical reason long-wave radio is still in use.
Step-by-step solution
1. Step 1
  Diffraction is most pronounced when the obstacle (or gap) is similar in size to the wavelength.
2. Step 2
  Radio: $\lambda = 1500\,\text{m} \gg 30\,\text{m}$ → the hill is small compared with $\lambda$ , so the radio wave diffracts strongly around it.
3. Step 3
  TV: $\lambda = 0.5\,\text{m} \ll 30\,\text{m}$ → very little diffraction → sharp 'shadow' behind the hill.
4. Step 4
  (b) Long-wave radio bends well around hills and buildings, giving wide coverage from a single transmitter without line-of-sight.
Answer
(a) The long-wave radio signal — its wavelength is comparable to or larger than the hill, so it diffracts strongly into the shadow region. (b) Diffraction gives wide-area coverage without line of sight.

Model Answers — General Properties of Waves

High-scoring sample answers for general properties of waves on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
State what is meant by the amplitude of a wave.
Model answer
The amplitude is the maximum displacement of a point on the wave from its rest (undisturbed) position.
Why this scores
One mark for 'maximum displacement from the rest position'. 'The height of the wave' is too loose — it must be measured from the centre line, not crest to trough.
Question 2
Paper 2/4 style2 marks
A wave has a frequency of $50\,\text{Hz}$ and a wavelength of $6.0\,\text{m}$ . Calculate its speed.
Model answer
$v = f\lambda = 50 \times 6.0 = \mathbf{300\,\text{m/s}}.$
Why this scores
One mark for substitution into $v = f\lambda$ , one for $300\,\text{m/s}$ with the unit.
Question 3
Paper 4 structured style3 marks
Explain the difference between a transverse wave and a longitudinal wave, giving one example of each.
Model answer
In a transverse wave the oscillations are perpendicular (at right angles) to the direction the wave travels — for example, light (or any electromagnetic wave), water waves, or seismic S-waves. In a longitudinal wave the oscillations are parallel to (along) the direction the wave travels, forming compressions and rarefactions — for example, sound (or seismic P-waves).
Why this scores
Three marks: transverse = oscillations perpendicular to travel (1); longitudinal = oscillations parallel to travel (1); a correct example of each (1). Calling sound transverse is the recurring error.
Question 4
Paper 4 structured style4 marks
A water wave of frequency $10\,\text{Hz}$ has a wavelength of $30\,\text{mm}$ in deep water. When it passes into shallow water its speed falls to $0.15\,\text{m/s}$ . (a) Calculate its speed in the deep water. (b) Calculate its wavelength in the shallow water.
Model answer
(a) In deep water $v_{1} = f\lambda_{1} = 10 \times 0.030 = \mathbf{0.30\,\text{m/s}}$ .

(b) The frequency does not change when a wave crosses a boundary (it is fixed by the source). So in shallow water

$\lambda_{2} = \dfrac{v_{2}}{f} = \dfrac{0.15}{10} = 0.015\,\text{m} = \mathbf{15\,\text{mm}}.$
Why this scores
Four marks: deep-water speed $0.30\,\text{m/s}$ (1); statement that frequency is unchanged (1); use of $\lambda = v/f$ (1); $15\,\text{mm}$ (1). Changing the frequency at the boundary is the classic error — only speed and wavelength change on refraction.
Question 5
Paper 4 explanation style5 marks
Plane water waves in a ripple tank pass through a gap in a barrier. (a) Describe what happens to the waves as they pass through the gap. (b) Explain how the spreading changes when the gap is made (i) wider and (ii) narrower, comparable to the wavelength.
Model answer
(a) The waves spread out (diffract) after passing through the gap; beyond the gap the wavefronts become curved at the edges instead of staying perfectly straight. The wavelength, frequency and speed are unchanged — only the shape/direction of the wavefronts changes.

(b)(i) When the gap is much wider than the wavelength, there is only a little diffraction — the waves carry on almost straight, with curving only at the edges. (ii) When the gap is made narrower, similar in size to the wavelength, the diffraction is much greater — the waves spread out into semicircular wavefronts filling the space beyond the gap. Diffraction is greatest when the gap width is about equal to the wavelength.
Why this scores
Five marks: waves spread out/diffract (1); wavelength/frequency/speed unchanged (1); wide gap → little spreading (1); narrow gap (≈ wavelength) → large spreading (1); maximum diffraction when gap ≈ wavelength (1).
Question 6
Paper 4 multi-part structured style6 marks
(a) Plane waves strike a straight barrier at $40\,\text{°}$ to the normal. State the angle of reflection and what happens to the speed, wavelength and frequency. (b) Plane waves then travel from deep into shallow water, slowing down. State what happens to the speed, wavelength, frequency and direction, and explain why the direction changes.
Model answer
(a) The angle of reflection equals the angle of incidence, so it is $40\,\text{°}$ from the normal. On reflection the speed, wavelength and frequency are all unchanged — only the direction of travel changes.

(b) Going from deep to shallow water, the speed decreases and (since $v = f\lambda$ and the frequency stays the same) the wavelength decreases. The direction changes — the waves bend towards the normal. This happens because one end of each wavefront enters the shallow water and slows down before the other end, so the wavefront pivots/changes direction (refraction).
Why this scores
Six marks: angle of reflection $40\,\text{°}$ (1); reflection leaves speed/wavelength/frequency unchanged (1); refraction — speed decreases (1); wavelength decreases while frequency unchanged (1); bends towards the normal (1); because one part of the wavefront slows before the rest (1).

Key Formulae — General Properties of Waves

The formulae you need to memorise for general properties of waves on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Wave equation
$v = f \lambda$
$v$
wave speed (m/s)
$f$
frequency (Hz)
$\lambda$
wavelength (m)
When to use
Any progressive wave.
Period - frequency
$T = \dfrac{1}{f}$
When to use
Convert between time for one cycle and number of cycles per second.

Key Definitions and Keywords — General Properties of Waves

Definitions to memorise and the exact keywords mark schemes credit for general properties of waves answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Wavelength
Examiner keyword
The shortest distance between two corresponding points on adjacent waves (e.g. crest to crest). Symbol $\lambda$ , unit metres (m).
Frequency
Examiner keyword
Number of complete oscillations per second. SI unit hertz (Hz).
Amplitude
Examiner keyword
Maximum displacement from the rest position.
Wavefront
Examiner keyword
A line (or surface) joining all points of a wave that are vibrating in step (in phase), e.g. a line of crests. Adjacent wavefronts are one wavelength apart.
Transverse wave
Examiner keyword
Wave in which oscillations are PERPENDICULAR to the direction of energy travel. Examples: electromagnetic radiation, water waves and seismic S-waves.
Longitudinal wave
Examiner keyword
Wave in which oscillations are PARALLEL to the direction of energy travel. Examples: sound waves and seismic P-waves.
Diffraction
Examiner keyword
Spreading of waves as they pass through a gap or around an obstacle. Most pronounced when gap ≈ wavelength.

Common Mistakes and Misconceptions — General Properties of Waves

The traps other students keep falling into on general properties of waves questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Saying sound is a transverse wave
0625/42 — recurring
Why it happens
Default to transverse mental image.
How to avoid it
Sound = LONGITUDINAL (compressions/rarefactions parallel to travel).
✕Using kHz or MHz directly without converting
Why it happens
Frequencies in real questions use prefixes.
How to avoid it
Convert to base Hz: $1\,\text{kHz} = 10^{3}\,\text{Hz}$ .
✕Saying gap size has no effect on diffraction
Why it happens
Confusing two diagrams in textbooks.
How to avoid it
Diffraction is greatest when gap ≈ wavelength.
✕Confusing oscillation distance with wavelength
Why it happens
Misreading a snapshot diagram.
How to avoid it
$\lambda$ is the spatial period (crest to crest), not the displacement of a particle.

General Properties of Waves — frequently asked questions

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

Cambridge IGCSE Physics (0625)

General Properties of Waves

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Short Notes - General Properties of Waves

Detailed Study Notes

Detailed notes on Waves for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

General Properties of Waves — Cambridge IGCSE 0625 Physics Extended (2026)

Transverse vs longitudinal, wavelength, frequency, period, $v = f\lambda$ , plus reflection, refraction and diffraction. The shared toolkit for sound, light and the EM spectrum.

At a glance

Transverse: oscillations PERPENDICULAR to direction of travel — EM radiation, water waves, seismic S-waves.
Longitudinal: oscillations PARALLEL to direction — sound waves and seismic P-waves.
Wavefront: a line joining points of a wave that are in step; adjacent wavefronts are one wavelength apart.
Wavelength $\lambda$ : distance for one full cycle ( $\text{m}$ ).
Frequency $f$ : cycles per second ( $\text{Hz}$ ). Period $T = 1/f$ .
Wave equation: $v = f\lambda$ .
Reflection / refraction / diffraction: angles equal; speed change at a boundary; spreading round gaps.

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

3.1 — Distinguish between transverse and longitudinal waves, using named examples (EM radiation, water and seismic S-waves as transverse; sound and seismic P-waves as longitudinal).
3.1 — Describe the features of a wave in terms of wavefront, wavelength, frequency, crest, trough and amplitude.
3.1 — Recall and use $v = f\lambda$ .
3.2 — Describe reflection, refraction and diffraction of waves.

Transverse and longitudinal waves

Transverse: peaks/troughs perpendicular to motion. Longitudinal: compressions/rarefactions along motion.

Seismic waves are produced by earthquakes and travel through the Earth. Remember: P for Parallel vibration (longitudinal) and S for Sideways vibration (transverse).

Both types share the features wavefront, wavelength, frequency, crest (peak), trough, amplitude and wave speed.

Drawing.

Transverse: a sine curve with crests (peaks) and troughs.
Longitudinal: a series of compressions (close particles) and rarefactions (spread particles).

Energy not matter. Waves transfer ENERGY without transferring matter. A floating cork bobs up and down as water waves pass — it doesn't travel with the wave.

Transverse waves oscillate across the direction of travel; longitudinal waves squeeze and stretch along it.

Transverse: oscillation $\perp$ travel — EM radiation, water, seismic S-waves.
Longitudinal: oscillation $\parallel$ travel — sound, seismic P-waves.
Wavefront: a line of in-step points; adjacent wavefronts one $\lambda$ apart.
Both transfer energy, not matter.

Wave parameters and the wave equation

$v = f\lambda$ . Memorise definitions of $\lambda, f, T$ .

Wavelength $\lambda$ : distance from one peak to the next (or one compression to the next). Units: $\text{m}$ .

Frequency $f$ : number of complete waves per second. Units: hertz ( $\text{Hz}$ ).

Period $T$ : time for one complete oscillation. $T = 1/f$ . Units: $\text{s}$ .

Amplitude $A$ : maximum displacement from rest position. Determines how much energy the wave carries.

Wave speed $v$ : how fast the wave travels. $\text{m/s}$ .

Wavelength λ is one full cycle (crest to crest); amplitude A is the maximum displacement from the rest position.

Wave equation. $v = f\lambda.$

Worked. A water wave has wavelength $0.5\,\text{m}$ and frequency $2\,\text{Hz}$ . Find speed.

$v = 2 \times 0.5 = 1\,\text{m/s}$ .

Worked. Sound in air at $340\,\text{m/s}$ , frequency $500\,\text{Hz}$ . Find wavelength.

$\lambda = v/f = 340/500 = 0.68\,\text{m}$ .

Tip. Always rearrange the equation to isolate the unknown — don't try to memorise three variants.

$v = f\lambda$ .
$T = 1/f$ .
Amplitude $\to$ energy carried.
Higher frequency $\to$ shorter wavelength (at fixed $v$ ).

Reflection and refraction

Reflection: angles equal. Refraction: speed change at boundary, direction bends.

Reflection. When a wave bounces off a surface.

Law of reflection: angle of incidence = angle of reflection (both measured from the NORMAL, not the surface).

Refraction. When a wave crosses into a different medium, its speed changes — and if it hits at an angle, its direction bends.

Going into a denser/slower medium → wave SLOWS down → bends TOWARD the normal.
Going into a less dense/faster medium → SPEEDS up → bends AWAY from normal.
Wavelength changes; frequency stays the same.

Worked qualitative. Light entering glass slows down → bends toward normal. Exiting back into air → speeds up → bends away from normal.

Tip. Always draw the normal (perpendicular) line at the boundary first. Both angles are measured from it.

Both angles are measured from the normal — never from the mirror surface.

Entering the slower medium the ray slows and bends toward the normal, so $r < i$; frequency is unchanged but wavelength shrinks.

Reflection: angles equal, measured from normal.
Refraction: speed AND direction change at boundary.
Slower medium: bend TOWARD normal.
Frequency stays constant; wavelength shrinks in slower medium.

Diffraction

Waves spread out as they pass through a gap or round an obstacle.

Diffraction. Waves bend or spread out when they pass through a gap or round an obstacle.

Amount of diffraction depends on the size of the gap relative to the wavelength:

Gap MUCH larger than $\lambda$ : little diffraction.
Gap $\approx \lambda$ : maximum diffraction.
Gap smaller than $\lambda$ : still spreads, but less wave gets through.

When the gap is about one wavelength wide, straight wavefronts emerge as circular ones — maximum diffraction.

Diffraction: waves spread round / through gaps.
Maximum when gap $\approx \lambda$ .
Sound diffracts well; light barely.
Demonstrated in ripple tanks.

Quick recap

Transverse (EM radiation, water, seismic S-waves) vs longitudinal (sound, seismic P-waves).
Wavefront = line of in-step points; adjacent wavefronts one $\lambda$ apart.
$v = f\lambda$ . $T = 1/f$ .
Reflection: angles equal (from normal).
Refraction: speed change → direction bend.
Diffraction: spreading; max when gap $\approx \lambda$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Transverse wave — oscillation perpendicular to direction of travel; e.g. EM radiation, water waves, seismic S-waves.
Longitudinal wave — oscillation parallel to direction of travel; e.g. sound waves, seismic P-waves.
Wavefront — a line joining points of a wave that are vibrating in step.
Wavelength — distance between two consecutive identical points on a wave.
Frequency — number of complete waves per second, $\text{Hz}$ .
Wave equation — $v = f\lambda$ .
Diffraction — spreading of waves around obstacles or through gaps.

How it’s examined

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (3.1-3.2); 0625/42 Oct/Nov 2024 — Q13 (ripple tank diffraction); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

Step-by-step worked examples — General Properties of Waves

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

Question type:

Question patterns to master — General Properties of Waves

Almost every general properties of waves exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

Two of $v$ , $f$ , $\lambda$ (or a period) are given and the third is wanted — find the speed / wavelength / frequency. One formula, one answer.

How to approach it

Use $v=f\lambda$ (or $f=1/T$ ), rearrange for the unknown, convert all quantities to SI base units, then substitute.

Common trap

Examiner reports flag unconverted prefixes — using kHz or MHz directly, or leaving $\lambda$ in mm. Convert to Hz and m before substituting, and always state the unit.

Identify & classify

Recognise it by

A list of waves (water, sound, light) with an instruction to classify them as transverse or longitudinal.

How to approach it

Compare the direction of oscillation with the direction of energy travel: perpendicular → transverse; parallel (compressions and rarefactions) → longitudinal.

Common trap

Examiner reports flag sound being called transverse. Sound is longitudinal — it travels as compressions and rarefactions along the direction of travel.

Graph or diagram

Recognise it by

A displacement–distance or displacement–time graph, a ripple-tank wavefront diagram, or an instruction to sketch diffraction or reflection.

How to approach it

Common trap

Examiner reports flag angles measured from the barrier instead of the normal, and confusing a particle's displacement with the wavelength on a snapshot graph.

Multi-step problem

Recognise it by

A wave crossing a boundary (refraction) where you must find a speed in one medium and a wavelength in another — two linked steps.

How to approach it

Apply $v=f\lambda$ in the first medium, carry the unchanged frequency across the boundary, then apply $v=f\lambda$ again in the second medium.

Common trap

Examiner reports flag candidates changing the frequency at a boundary. Frequency is fixed by the source — only $v$ and $\lambda$ change on refraction.

Show that / prove

Recognise it by

The stem says explain why or state which … and why — typically comparing how strongly two wavelengths diffract around an obstacle.

How to approach it

Argue from the rule that diffraction is greatest when the obstacle or gap is comparable in size to the wavelength, then apply it to each case and conclude.

Common trap

Claiming gap or obstacle size has no effect on diffraction — the amount of spreading depends strongly on the wavelength-to-gap ratio.

1Use the wave equation
CoreDirect calculation• v=fλ
Question
A water wave has frequency $4\,\text{Hz}$ and wavelength $0.5\,\text{m}$ . Find its speed.
Step-by-step solution
1. Step 1
  $v = f\lambda$ .
  $v = 4 \times 0.5 = 2\,\text{m/s}$
Answer
$2\,\text{m/s}$
2Period and frequency
CoreDirect calculation• period
Question
A wave has period $0.02\,\text{s}$ . Find its frequency.
Step-by-step solution
1. Step 1
  $f = 1/T$ .
  $f = \dfrac{1}{0.02} = 50\,\text{Hz}$
Answer
$50\,\text{Hz}$
3Distinguish transverse and longitudinal
CoreIdentify & classify• wave types
Question
Classify: water waves on a pond, sound in air, EM waves.
Step-by-step solution
1. Step 1
  Transverse: oscillation perpendicular to wave direction.
2. Step 2
  Longitudinal: oscillation parallel to wave direction (compressions and rarefactions).
Answer
Water and EM: transverse. Sound: longitudinal.
4Seismic P-waves and S-waves — transverse or longitudinal?
Core• wave types, seismic
Question
An earthquake produces two types of seismic wave: P-waves (primary) and S-waves (secondary). (a) State which of these can be modelled as transverse and which as longitudinal. (b) For each, describe the direction of vibration relative to the direction the wave travels.
Step-by-step solution
1. Step 1
  Recall the named examples: sound and seismic P-waves are modelled as longitudinal; electromagnetic radiation, water waves and seismic S-waves are modelled as transverse.
2. Step 2
  (a) Seismic P-waves are longitudinal; seismic S-waves are transverse.
3. Step 3
  (b) In a P-wave the rock particles vibrate back and forth PARALLEL to the direction the wave travels (compressions and rarefactions). In an S-wave the rock particles vibrate at RIGHT ANGLES (perpendicular) to the direction the wave travels.
Answer
(a) P-waves: longitudinal. S-waves: transverse. (b) P-wave vibrations are parallel to the direction of travel; S-wave vibrations are perpendicular to it.
Examiner tip
P-waves and S-waves are the named examples in the 2026 syllabus §3.1 — remember 'P for Parallel (longitudinal)' and 'S for Sideways (transverse)'.
5Describing a wavefront
Core• wavefront, wave types
Question
A ripple tank produces straight (plane) water waves. (a) Explain what is meant by a wavefront. (b) State how the distance between two adjacent wavefronts is related to the wavelength.
Step-by-step solution
1. Step 1
  (a) A wavefront is a line (or surface) joining all the points on a wave that are vibrating exactly in step — for example, a line joining all the crests of one ripple.
2. Step 2
  (b) Adjacent wavefronts are one wavelength apart, because successive crests are separated by one wavelength.
Answer
(a) A wavefront joins all points of a wave that are in phase (e.g. a line of crests). (b) The distance between two adjacent wavefronts equals one wavelength.
6Diffraction through a gap
ExtendedGraph or diagram• Adapted from 0625/42 May/Jun 2024 Q12• diffraction
Question
Sketch how diffraction changes when the gap is similar to the wavelength versus much larger than the wavelength.
Step-by-step solution
1. Step 1
  Gap ≈ wavelength → significant spreading (large diffraction).
2. Step 2
  Gap ≫ wavelength → almost no spreading (waves continue almost straight).
Answer
Maximum diffraction occurs when the gap width is similar to the wavelength.
7Find the wavelength from speed and frequency
ExtendedDirect calculation• Adapted from 0625/42 Oct/Nov 2023 Q13• v=fλ
Question
A sound wave in air travels at $340\,\text{m/s}$ with a frequency of $1700\,\text{Hz}$ . Find the wavelength.
Step-by-step solution
1. Step 1
  $v = f\lambda \Rightarrow \lambda = v/f$ .
  $\lambda = \dfrac{340}{1700} = 0.20\,\text{m}$
Answer
$\lambda = 0.20\,\text{m} = 20\,\text{cm}$
Examiner tip
The examiner report flags candidates often quote the answer in metres without unit conversion checks (e.g. mm vs m). Always state the unit explicitly.
8Read amplitude and wavelength from a graph
ExtendedGraph or diagram• graph, transverse
Question
A transverse wave on a string is drawn on a displacement-distance graph. The y-axis ranges $-3\,\text{cm}$ to $+3\,\text{cm}$ ; the wave shows two complete cycles between $x = 0\,\text{m}$ and $x = 1.0\,\text{m}$ . State the amplitude and wavelength.
Step-by-step solution
1. Step 1
  Amplitude = maximum displacement from rest.
  $A = 3\,\text{cm} = 0.03\,\text{m}$
2. Step 2
  Two cycles in $1.0\,\text{m}$ → one cycle in $0.50\,\text{m}$ .
  $\lambda = 0.50\,\text{m}$
Answer
$A = 0.03\,\text{m}$ ; $\lambda = 0.50\,\text{m}$
9Read period and frequency from a graph
ExtendedGraph or diagram• graph, period
Question
A displacement-time graph for a wave shows three complete oscillations between $t = 0$ and $t = 0.12\,\text{s}$ . Find the period and frequency.
Step-by-step solution
1. Step 1
  Period = time for one complete oscillation.
  $T = \dfrac{0.12}{3} = 0.04\,\text{s}$
2. Step 2
  $f = 1/T$ .
  $f = \dfrac{1}{0.04} = 25\,\text{Hz}$
Answer
$T = 0.04\,\text{s}$ ; $f = 25\,\text{Hz}$
10Reflection of plane waves at a barrier
ExtendedGraph or diagram• reflection, ripple
Question
Plane water waves in a ripple tank approach a straight barrier at $30\degree$ to the normal. State (a) the angle of reflection, (b) what happens to the wavelength, frequency and speed of the reflected wave.
Step-by-step solution
1. Step 1
  Angle of reflection = angle of incidence, both measured from the normal.
  $\theta_{r} = 30\degree$
2. Step 2
  Reflection does NOT change wavelength, frequency or speed — only direction.
Answer
(a) $30\degree$ from the normal. (b) Wavelength, frequency and speed are all unchanged.
Examiner tip
The examiner report flags candidates often measure the angle from the barrier itself. Always measure from the normal — the perpendicular to the barrier.
11Refraction of a wave at a boundary
ChallengeMulti-step problem• Adapted from 0625/42 May/Jun 2024 Q12• refraction, wave
Question
A water wave of frequency $5.0\,\text{Hz}$ has wavelength $40\,\text{mm}$ in deep water. When it crosses into shallow water its speed falls to $0.10\,\text{m/s}$ . Find (a) the speed in deep water, (b) the wavelength in shallow water.
Step-by-step solution
1. Step 1
  Speed in deep water.
  $v_{1} = f\lambda_{1} = 5.0 \times 0.040 = 0.20\,\text{m/s}$
2. Step 2
  Frequency is unchanged on crossing a boundary; only speed and wavelength change.
  $\lambda_{2} = \dfrac{v_{2}}{f} = \dfrac{0.10}{5.0}$
3. Step 3
  Compute.
  $\lambda_{2} = 0.020\,\text{m} = 20\,\text{mm}$
Answer
(a) $v_{1} = 0.20\,\text{m/s}$ . (b) $\lambda_{2} = 20\,\text{mm}$ .
Examiner tip
The examiner report flags candidates often change frequency at the boundary. Frequency is fixed by the source — only $v$ and $\lambda$ change on refraction.
12A* — Compare diffraction of long and short wavelengths
ChallengeShow that / prove• diffraction, synoptic
Question
A long-wave radio station broadcasts at $\lambda \approx 1500\,\text{m}$ and a TV transmitter broadcasts at $\lambda \approx 0.5\,\text{m}$ . Both encounter a hill $30\,\text{m}$ wide blocking the line of sight to a house behind it. (a) State which signal is received more strongly behind the hill and why. (b) Use this to suggest one practical reason long-wave radio is still in use.
Step-by-step solution
1. Step 1
  Diffraction is most pronounced when the obstacle (or gap) is similar in size to the wavelength.
2. Step 2
  Radio: $\lambda = 1500\,\text{m} \gg 30\,\text{m}$ → the hill is small compared with $\lambda$ , so the radio wave diffracts strongly around it.
3. Step 3
  TV: $\lambda = 0.5\,\text{m} \ll 30\,\text{m}$ → very little diffraction → sharp 'shadow' behind the hill.
4. Step 4
  (b) Long-wave radio bends well around hills and buildings, giving wide coverage from a single transmitter without line-of-sight.
Answer
(a) The long-wave radio signal — its wavelength is comparable to or larger than the hill, so it diffracts strongly into the shadow region. (b) Diffraction gives wide-area coverage without line of sight.

Model Answers — General Properties of Waves

High-scoring sample answers for general properties of waves on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
State what is meant by the amplitude of a wave.
Model answer
The amplitude is the maximum displacement of a point on the wave from its rest (undisturbed) position.
Why this scores
One mark for 'maximum displacement from the rest position'. 'The height of the wave' is too loose — it must be measured from the centre line, not crest to trough.
Question 2
Paper 2/4 style2 marks
A wave has a frequency of $50\,\text{Hz}$ and a wavelength of $6.0\,\text{m}$ . Calculate its speed.
Model answer
$v = f\lambda = 50 \times 6.0 = \mathbf{300\,\text{m/s}}.$
Why this scores
One mark for substitution into $v = f\lambda$ , one for $300\,\text{m/s}$ with the unit.
Question 3
Paper 4 structured style3 marks
Explain the difference between a transverse wave and a longitudinal wave, giving one example of each.
Model answer
In a transverse wave the oscillations are perpendicular (at right angles) to the direction the wave travels — for example, light (or any electromagnetic wave), water waves, or seismic S-waves. In a longitudinal wave the oscillations are parallel to (along) the direction the wave travels, forming compressions and rarefactions — for example, sound (or seismic P-waves).
Why this scores
Three marks: transverse = oscillations perpendicular to travel (1); longitudinal = oscillations parallel to travel (1); a correct example of each (1). Calling sound transverse is the recurring error.
Question 4
Paper 4 structured style4 marks
A water wave of frequency $10\,\text{Hz}$ has a wavelength of $30\,\text{mm}$ in deep water. When it passes into shallow water its speed falls to $0.15\,\text{m/s}$ . (a) Calculate its speed in the deep water. (b) Calculate its wavelength in the shallow water.
Model answer
(a) In deep water $v_{1} = f\lambda_{1} = 10 \times 0.030 = \mathbf{0.30\,\text{m/s}}$ .

(b) The frequency does not change when a wave crosses a boundary (it is fixed by the source). So in shallow water

$\lambda_{2} = \dfrac{v_{2}}{f} = \dfrac{0.15}{10} = 0.015\,\text{m} = \mathbf{15\,\text{mm}}.$
Why this scores
Four marks: deep-water speed $0.30\,\text{m/s}$ (1); statement that frequency is unchanged (1); use of $\lambda = v/f$ (1); $15\,\text{mm}$ (1). Changing the frequency at the boundary is the classic error — only speed and wavelength change on refraction.
Question 5
Paper 4 explanation style5 marks
Plane water waves in a ripple tank pass through a gap in a barrier. (a) Describe what happens to the waves as they pass through the gap. (b) Explain how the spreading changes when the gap is made (i) wider and (ii) narrower, comparable to the wavelength.
Model answer
(a) The waves spread out (diffract) after passing through the gap; beyond the gap the wavefronts become curved at the edges instead of staying perfectly straight. The wavelength, frequency and speed are unchanged — only the shape/direction of the wavefronts changes.

(b)(i) When the gap is much wider than the wavelength, there is only a little diffraction — the waves carry on almost straight, with curving only at the edges. (ii) When the gap is made narrower, similar in size to the wavelength, the diffraction is much greater — the waves spread out into semicircular wavefronts filling the space beyond the gap. Diffraction is greatest when the gap width is about equal to the wavelength.
Why this scores
Five marks: waves spread out/diffract (1); wavelength/frequency/speed unchanged (1); wide gap → little spreading (1); narrow gap (≈ wavelength) → large spreading (1); maximum diffraction when gap ≈ wavelength (1).
Question 6
Paper 4 multi-part structured style6 marks
(a) Plane waves strike a straight barrier at $40\,\text{°}$ to the normal. State the angle of reflection and what happens to the speed, wavelength and frequency. (b) Plane waves then travel from deep into shallow water, slowing down. State what happens to the speed, wavelength, frequency and direction, and explain why the direction changes.
Model answer
(a) The angle of reflection equals the angle of incidence, so it is $40\,\text{°}$ from the normal. On reflection the speed, wavelength and frequency are all unchanged — only the direction of travel changes.

(b) Going from deep to shallow water, the speed decreases and (since $v = f\lambda$ and the frequency stays the same) the wavelength decreases. The direction changes — the waves bend towards the normal. This happens because one end of each wavefront enters the shallow water and slows down before the other end, so the wavefront pivots/changes direction (refraction).
Why this scores
Six marks: angle of reflection $40\,\text{°}$ (1); reflection leaves speed/wavelength/frequency unchanged (1); refraction — speed decreases (1); wavelength decreases while frequency unchanged (1); bends towards the normal (1); because one part of the wavefront slows before the rest (1).

Key Formulae — General Properties of Waves

The formulae you need to memorise for general properties of waves on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Wave equation
$v = f \lambda$
$v$
wave speed (m/s)
$f$
frequency (Hz)
$\lambda$
wavelength (m)
When to use
Any progressive wave.
Period - frequency
$T = \dfrac{1}{f}$
When to use
Convert between time for one cycle and number of cycles per second.

Key Definitions and Keywords — General Properties of Waves

Definitions to memorise and the exact keywords mark schemes credit for general properties of waves answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Wavelength
Examiner keyword
The shortest distance between two corresponding points on adjacent waves (e.g. crest to crest). Symbol $\lambda$ , unit metres (m).
Frequency
Examiner keyword
Number of complete oscillations per second. SI unit hertz (Hz).
Amplitude
Examiner keyword
Maximum displacement from the rest position.
Wavefront
Examiner keyword
A line (or surface) joining all points of a wave that are vibrating in step (in phase), e.g. a line of crests. Adjacent wavefronts are one wavelength apart.
Transverse wave
Examiner keyword
Wave in which oscillations are PERPENDICULAR to the direction of energy travel. Examples: electromagnetic radiation, water waves and seismic S-waves.
Longitudinal wave
Examiner keyword
Wave in which oscillations are PARALLEL to the direction of energy travel. Examples: sound waves and seismic P-waves.
Diffraction
Examiner keyword
Spreading of waves as they pass through a gap or around an obstacle. Most pronounced when gap ≈ wavelength.

Common Mistakes and Misconceptions — General Properties of Waves

The traps other students keep falling into on general properties of waves questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Saying sound is a transverse wave
0625/42 — recurring
Why it happens
Default to transverse mental image.
How to avoid it
Sound = LONGITUDINAL (compressions/rarefactions parallel to travel).
✕Using kHz or MHz directly without converting
Why it happens
Frequencies in real questions use prefixes.
How to avoid it
Convert to base Hz: $1\,\text{kHz} = 10^{3}\,\text{Hz}$ .
✕Saying gap size has no effect on diffraction
Why it happens
Confusing two diagrams in textbooks.
How to avoid it
Diffraction is greatest when gap ≈ wavelength.
✕Confusing oscillation distance with wavelength
Why it happens
Misreading a snapshot diagram.
How to avoid it
$\lambda$ is the spatial period (crest to crest), not the displacement of a particle.

General Properties of Waves — frequently asked questions

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

General Properties of Waves

Short Notes - General Properties of Waves

Detailed Study Notes

At a glance

What you’ll learn

Transverse and longitudinal waves

Wave parameters and the wave equation

Reflection and refraction

Diffraction

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — General Properties of Waves

1Use the wave equation

2Period and frequency

3Distinguish transverse and longitudinal

4Seismic P-waves and S-waves — transverse or longitudinal?

5Describing a wavefront

6Diffraction through a gap

7Find the wavelength from speed and frequency

8Read amplitude and wavelength from a graph

9Read period and frequency from a graph

10Reflection of plane waves at a barrier

11Refraction of a wave at a boundary

12A* — Compare diffraction of long and short wavelengths

Wave equation

Period - frequency

Wavelength

Frequency

Amplitude

Wavefront

Transverse wave

Longitudinal wave

Diffraction

✕Saying sound is a transverse wave

✕Using kHz or MHz directly without converting

✕Saying gap size has no effect on diffraction

✕Confusing oscillation distance with wavelength

General Properties of Waves — frequently asked questions

General Properties of Waves

Short Notes - General Properties of Waves

Detailed Study Notes

At a glance

What you’ll learn

Transverse and longitudinal waves

Wave parameters and the wave equation

Reflection and refraction

Diffraction

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — General Properties of Waves

1Use the wave equation

2Period and frequency

3Distinguish transverse and longitudinal

4Seismic P-waves and S-waves — transverse or longitudinal?

5Describing a wavefront

6Diffraction through a gap

7Find the wavelength from speed and frequency

8Read amplitude and wavelength from a graph

9Read period and frequency from a graph

10Reflection of plane waves at a barrier

11Refraction of a wave at a boundary

12A* — Compare diffraction of long and short wavelengths

Wave equation

Period - frequency

Wavelength

Frequency

Amplitude

Wavefront

Transverse wave