Why is "Reading the wavelength off a displacement–time graph" a common mistake in Wave model?

Why it happens: The two graphs look identical (both are sine curves), so students grab the crest-to-crest gap without checking the horizontal axis. How to avoid it: Check the horizontal axis first. On a **time** axis the crest-to-crest gap is the **period T** (then f = 1/T); only a **distance** axis gives the **wavelength λ**. Source: IB Physics Theme C subject reports

Why is "Confusing frequency with period" a common mistake in Wave model?

Why it happens: Both describe how often the wave oscillates, and students forget which is time and which is a rate. How to avoid it: Period T is a **time** (seconds) for one wave; frequency f is a **rate** (waves per second, Hz). They are reciprocals: T = 1/f. Convert a period to a frequency before using v = fλ. Source: IB Physics Theme C subject reports

Why is "Thinking the particles travel along with the wave" a common mistake in Wave model?

Why it happens: The wave visibly moves forward, so students assume the medium moves with it. How to avoid it: State clearly that a wave transfers **energy, not matter**. Each particle only **oscillates about a fixed rest position**; it does not move along with the wave (think of a bobbing duck). Source: IB Physics Theme C subject reports

Why is "Mixing up transverse and longitudinal examples" a common mistake in Wave model?

Why it happens: Students memorise the definitions but not which everyday waves belong to each class — sound is especially often mislabelled as transverse. How to avoid it: Anchor the examples: **sound = longitudinal**; **light and all EM waves = transverse**; waves on a string and water surface waves are transverse. Test each by asking which way the particles oscillate relative to the energy transfer. Source: IB Physics Theme C subject reports

Why is "Thinking different EM waves travel at different speeds" a common mistake in Wave model?

Why it happens: Because gamma rays are 'higher energy', students assume they are also faster than radio waves. How to avoid it: Remember that **all** EM waves travel at the **same speed c = 3.0×10⁸ m s⁻¹ in a vacuum**. They differ only in frequency and wavelength (linked by c = fλ), never in speed. Source: IB Physics Theme C subject reports

Why is "Unit and prefix slips (nm, MHz, GHz) in wave calculations" a common mistake in Wave model?

Why it happens: Wave problems mix very large and very small numbers, and students forget to convert prefixes to SI before substituting. How to avoid it: Always convert to base SI first: 1 nm = 10⁻⁹ m, 1 MHz = 10⁶ Hz, 1 GHz = 10⁹ Hz. Substitute the converted values into v = fλ and quote the final answer with its unit. Source: IB Physics Theme C subject reports

IB DP Physics (100452)

Wave model

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Detailed Study Notes

Detailed notes on Wave behaviour for IB DP Physics, covering key concepts, explanations, examples, and exam-focused revision points.

C.2 Wave model — IB Physics SL Study Notes (Theme C: Wave Behaviour)

What a wave really is — a transfer of energy without a net transfer of matter — and the language that describes it: transverse vs longitudinal waves, wavelength, frequency, period, amplitude and speed, the wave equation v = fλ, reading displacement–distance and displacement–time graphs, wavefronts and rays, and the electromagnetic spectrum.

At a glance

A wave transfers energy without a net transfer of matter — the particles of the medium just oscillate about a fixed point.
Transverse waves oscillate perpendicular to the energy transfer (e.g. light, waves on a string); longitudinal waves oscillate parallel to it, with compressions and rarefactions (e.g. sound).
The wave equation ties everything together: $v = f\lambda$ .
Period and frequency are reciprocals: $T = \dfrac{1}{f}$ . A frequency of $f$ hertz means $f$ complete oscillations every second.
A displacement–distance graph shows the wavelength $\lambda$ ; a displacement–time graph shows the period $T$ . Never read $\lambda$ off a time graph.
All electromagnetic waves are transverse and travel at $c = 3.0\times10^{8}\,\text{m s}^{-1}$ in a vacuum (order of increasing $f$ : radio → microwave → IR → visible → UV → X-ray → gamma).
Sound is a longitudinal mechanical wave: it needs a medium and cannot travel through a vacuum.
Amplitude is the maximum displacement from the rest position; it can be read from either graph and (for the same wave) is the same on both.

What you’ll learn

Mapped to the 100452 subject guide (2025-onwards).

Describe a wave as the transfer of energy without a net transfer of matter, and distinguish transverse from longitudinal waves with correct examples.
Define and use the wave quantities wavelength, frequency, period, amplitude and wave speed, with their SI units.
Apply the wave equation $v = f\lambda$ together with the relation $T = 1/f$ to solve problems, including for electromagnetic waves.
Interpret displacement–distance and displacement–time graphs, extracting wavelength, period and amplitude, and explain the crucial difference between the two graph types.
Describe wavefronts and rays, recall the ordering of the electromagnetic spectrum, and explain that all EM waves travel at c in a vacuum while sound is a longitudinal mechanical wave that requires a medium.

What a wave is: energy on the move, matter staying put

A wave carries energy through a medium while the particles only oscillate about fixed points.

A wave is a disturbance that travels through space or a medium and transfers energy from one place to another without a net transfer of matter. This one sentence is the heart of C.2 — and it earns marks in almost every wave question.

Picture a Mexican wave in a stadium: the wave races round the ground, but each spectator only stands up and sits back down — nobody runs around the stadium with it. In the same way, when a water wave passes, a floating duck bobs up and down about a fixed point; it does not travel across the lake with the wave. The energy moves; the water (and the duck) does not.

Two great families of wave

Feature	Transverse wave	Longitudinal wave
Oscillation direction	Perpendicular to energy transfer	Parallel to energy transfer
Shape	crests and troughs	compressions and rarefactions
Examples	light & all EM waves, waves on a string, water surface waves	sound, a compression wave on a slinky, seismic P-waves
Needs a medium?	EM waves do not; string/water do	Yes — always a mechanical wave

In a transverse wave the particles move at right angles to the direction the energy travels. A wave on a rope shaken up-and-down travels along the rope while each bit of rope moves up and down.
In a longitudinal wave the particles move back and forth along the same line the energy travels. Regions where particles bunch together are compressions; regions where they spread out are rarefactions.

In a transverse wave the particle oscillation is at right angles to the energy transfer; in a longitudinal wave it is along the same line, producing compressions (particles close together) and rarefactions (particles spread out).

Mechanical vs electromagnetic. A mechanical wave (sound, water, string, seismic) needs a medium to travel through — it is the medium's particles that oscillate. An electromagnetic wave (light, radio, X-rays) is an oscillation of electric and magnetic fields and needs no medium, which is why sunlight reaches us across the vacuum of space.

A wave transfers energy without a net transfer of matter; particles oscillate about fixed points.
Transverse: oscillation perpendicular to propagation (light, string, water surface).
Longitudinal: oscillation parallel to propagation, with compressions and rarefactions (sound).

See the full worked example for wave model →

The five wave quantities you must know

Wavelength, frequency, period, amplitude and speed — each with its own symbol, meaning and unit.

Every wave is described by the same small set of quantities. Learn the symbol, the meaning and the unit for each — examiners test all three.

Quantity	Symbol	Meaning	SI unit
Wavelength	$\lambda$	Distance between two adjacent points in phase (e.g. crest to next crest)	m
Frequency	$f$	Number of complete oscillations (waves) passing a point per second	Hz (s⁻¹)
Period	$T$	Time for one complete oscillation	s
Amplitude	$A$	Maximum displacement of a particle from its rest position	m
Wave speed	$v$	Speed at which the wave (energy) travels	m s⁻¹

A few things students slip on:

Amplitude is measured from the middle (rest) line to a crest — not from a trough to a crest. Crest-to-trough is twice the amplitude.
"In phase" means two points moving in exactly the same way at the same time (e.g. two crests, or two troughs). Wavelength is the shortest distance between two such points.
Frequency and period are reciprocals: $T = \dfrac{1}{f}$ and $f = \dfrac{1}{T}$ . If a wave has a frequency of $50\,\text{Hz}$ , its period is $T = 1/50 = 0.020\,\text{s}$ .
Watch the prefixes: $1\,\text{kHz}=10^{3}\,\text{Hz}$ , $1\,\text{MHz}=10^{6}\,\text{Hz}$ , $1\,\text{GHz}=10^{9}\,\text{Hz}$ ; $1\,\text{nm}=10^{-9}\,\text{m}$ . A great many lost marks are simply unit-conversion slips.

Frequency is set by the source. When a wave passes from one medium into another its speed and wavelength can change, but its frequency stays the same because it is fixed by whatever is making the wave oscillate.

λ = wavelength (m), f = frequency (Hz), T = period (s), A = amplitude (m), v = speed (m s⁻¹).
Amplitude is measured from the rest position, not crest to trough.
T = 1/f; frequency is fixed by the source.

See the full worked example for wave model →

Deriving the wave equation v = fλ from first principles

In one period the wave advances exactly one wavelength — so speed = distance ÷ time gives v = fλ.

You are not asked to memorise a derivation for the exam, but seeing where $v = f\lambda$ comes from makes it impossible to misuse — and IB rewards students who reason rather than pattern-match.

Start from the everyday definition of speed: $v = \frac{\text{distance travelled}}{\text{time taken}}$

Now watch one full cycle of the wave. In a time of exactly one period $T$ , the wave pattern advances forward by exactly one wavelength $\lambda$ (the next crest arrives where the last crest was). So over that one cycle:

distance travelled $= \lambda$
time taken $= T$

Substituting into the definition of speed: $v = \frac{\lambda}{T}$

Finally, because period and frequency are reciprocals, $\dfrac{1}{T} = f$ , so: $v = \frac{\lambda}{T} = \lambda \times \frac{1}{T} = \lambda f \quad\Rightarrow\quad \boxed{v = f\lambda}$

That is the wave equation. It says: the faster the wave, or the more waves squeezed past each second, the more ground is covered — with wavelength as the length of each wave.

Between the solid and dashed snapshots, exactly one period T has elapsed and every crest has advanced by one wavelength λ. Distance ÷ time gives v = λ/T = fλ.

Why this matters for a grade 9: if you ever forget the formula, recover it in seconds — "in one period the wave moves one wavelength" gives $v = \lambda/T$ , and $1/T = f$ finishes the job. And because $f$ is fixed by the source, the equation immediately tells you that if a wave slows down on entering a new medium, its wavelength must shrink in proportion.

Speed = distance ÷ time; over one cycle the wave covers λ in a time T.
So v = λ/T, and since 1/T = f, v = fλ.
Because f is fixed by the source, a change in v forces a matching change in λ.

See the full worked example for wave model →

The two graphs: displacement–distance vs displacement–time

A distance graph reveals the wavelength; a time graph reveals the period. They look identical — read the axis.

Two graphs describe a wave, and they look almost the same — a sine curve. The only reliable way to tell them apart is to read the horizontal axis, and getting this wrong is the single most common wave error in the exam.

Displacement–distance graph (horizontal axis = distance / position along the wave)

This is a snapshot of the whole wave frozen at one instant, like a photograph.
The horizontal distance between two adjacent crests is the wavelength $\lambda$ .
The height from the middle line to a crest is the amplitude $A$ .
✗ You cannot read the period or frequency from this graph.

Displacement–time graph (horizontal axis = time)

This follows one single particle as time passes, like filming one duck bob up and down.
The horizontal time between two adjacent crests is the period $T$ — and then $f = 1/T$ .
The height from the middle line to a crest is again the amplitude $A$ .
✗ You cannot read the wavelength from this graph.

Two identical-looking curves. On the distance-axis graph the gap between crests is the wavelength λ; on the time-axis graph it is the period T. Amplitude A is read the same way on both.

The killer exam move: if you have both graphs for the same wave, read $\lambda$ from the distance graph and $T$ from the time graph, then combine them: $v = \frac{\lambda}{T} = f\lambda$ This is a favourite Paper 2 question because it forces you to use both graphs correctly.

Displacement–distance graph = snapshot of the whole wave → read the wavelength λ.
Displacement–time graph = one particle over time → read the period T (then f = 1/T).
Amplitude is read from either; with both graphs, v = λ/T.

See the full worked example for wave model →

Wavefronts, rays and the electromagnetic spectrum

Wavefronts and rays picture a wave in 2D; the EM spectrum lines up light of every wavelength — all at speed c.

Wavefronts and rays are two ways of drawing a wave spreading through space.

A wavefront is a line (or surface) joining points that are all in phase — for example, the line along the top of each crest. Adjacent wavefronts are one wavelength apart.
A ray is an arrow drawn at right angles to the wavefronts, showing the direction the wave (energy) travels.

Far from a small source, the wavefronts are almost straight parallel lines (plane waves); close to a point source they are circles (or spheres) expanding outwards.

Wavefronts join points in phase and are one wavelength apart; rays are drawn perpendicular to them, pointing in the direction the energy travels. Point sources give circular wavefronts; distant sources give plane wavefronts.

The electromagnetic (EM) spectrum

Light is just one member of a huge family of electromagnetic waves. They are all transverse, and — crucially — all travel at the same speed $c = 3.0\times10^{8}\,\text{m s}^{-1}$ in a vacuum. They differ only in frequency and wavelength (related by $c = f\lambda$ ). Learn the order:

Order of the EM spectrum by increasing frequency (decreasing wavelength): radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. Visible light runs red (longest λ) to violet (shortest λ). Every one of them travels at c in a vacuum.

A useful memory hook (longest wavelength → shortest): "Radio Microwaves Injure Various Unlucky X-ray Gamers" (Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma). Within visible light, red has the longest wavelength and violet the shortest.

Wavefronts join points in phase (one wavelength apart); rays are perpendicular arrows showing direction of travel.
EM spectrum order (increasing f): radio, microwave, IR, visible, UV, X-ray, gamma.
All EM waves are transverse and travel at c = 3.0×10⁸ m s⁻¹ in a vacuum.

See the full worked example for wave model →

Sound: a longitudinal mechanical wave

Sound is compressions and rarefactions travelling through a medium — it cannot cross a vacuum.

Sound is the most important example of a longitudinal mechanical wave. A vibrating source (a loudspeaker cone, a guitar string, your vocal cords) pushes on the air:

as it moves forward it squashes the air particles together → a compression (high pressure);
as it moves back it lets them spread apart → a rarefaction (low pressure).

These compressions and rarefactions travel outwards as a wave. The air particles themselves only oscillate back and forth about their rest positions along the direction of travel — they do not stream away from the loudspeaker. Once again: energy travels, matter does not.

Sound needs a medium. Because it is the particles of the medium that carry the disturbance, sound cannot travel through a vacuum. This is the classic "bell in a bell jar" demonstration: as the air is pumped out, the ringing bell grows silent even though you can still see it vibrating. In space, no one can hear anything — there are no particles to compress. (Light, being an EM wave, crosses the same vacuum with ease.)

The wave equation still applies to sound: $v = f\lambda$ . In air at room temperature the speed of sound is roughly $v \approx 340\,\text{m s}^{-1}$ — about a million times slower than light, which is why you see lightning well before you hear the thunder. Sound travels faster in liquids and solids than in gases, because the particles are closer together and pass the disturbance on more quickly.

A worked micro-example (do it in your head): a tuning fork of frequency $170\,\text{Hz}$ produces sound at $340\,\text{m s}^{-1}$ . Its wavelength is $\lambda = \frac{v}{f} = \frac{340}{170} = 2.0\,\text{m}.$ Notice the structure: the same $v = f\lambda$ that governs light also governs sound — only the numbers change.

Sound is a longitudinal mechanical wave: compressions and rarefactions.
It needs a medium — it cannot travel through a vacuum (bell-jar demo).
v ≈ 340 m s⁻¹ in air; faster in liquids and solids; still obeys v = fλ.

See the full worked example for wave model →

Quick recap

A wave transfers energy without a net transfer of matter; the particles oscillate about fixed points.
Transverse: oscillation perpendicular to propagation (EM waves, string, water); longitudinal: oscillation parallel, with compressions and rarefactions (sound).
The wave equation v = fλ and the relation T = 1/f connect all the wave quantities.
A displacement–distance graph gives the wavelength λ; a displacement–time graph gives the period T; both show the amplitude A.
All EM waves are transverse and travel at c = 3.0×10⁸ m s⁻¹ in a vacuum, in the order radio → microwave → IR → visible → UV → X-ray → gamma.
Sound is a longitudinal mechanical wave that needs a medium; it cannot travel through a vacuum.

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

A wave transfers energy without a net transfer of matter.
Transverse: oscillation perpendicular to propagation (light, string, water surface).
Longitudinal: oscillation parallel to propagation, with compressions and rarefactions (sound).
Wave equation: v = fλ.
Period–frequency: T = 1/f (so f = 1/T).
Displacement–distance graph → wavelength λ; displacement–time graph → period T.
All EM waves travel at c = 3.0×10⁸ m s⁻¹ in a vacuum (radio → microwave → IR → visible → UV → X-ray → gamma).
Sound is a longitudinal mechanical wave and cannot travel through a vacuum.

How it’s examined

The wave model underpins the whole of Theme C and reappears in optics, standing waves and the Doppler effect. Paper 1A (MCQ): classifying a wave as transverse or longitudinal, matching a described wave to a displacement–distance or displacement–time graph, one-step $v=f\lambda$ calculations (numbers chosen to be calculator-free), and ordering the EM spectrum. Paper 1B (data-based): reading the wavelength off a distance graph and the period off a time graph, then combining them to find speed or frequency, with marks for stating which quantity each graph gives. Paper 2: structured questions that provide both graphs and ask candidates to determine $\lambda$ , $T$ , $f$ , $A$ and $v$ in turn, plus 'describe/explain' questions on why sound cannot travel through a vacuum and why all EM waves share the same speed $c$ . Command terms: state, identify, determine, calculate, describe, explain, sketch. Examiner reports repeatedly flag: reading a wavelength off a time graph, confusing frequency with period, claiming particles travel with the wave, mixing up transverse and longitudinal examples, forgetting that all EM waves travel at $c$ , and prefix/unit slips (nm, MHz, GHz). Always quote the unit and show the substitution — method marks survive an arithmetic slip.

Sources: IB Diploma Programme Physics Guide (first assessment 2025) — Theme C: Wave behaviour (C.2 Wave model); IB Physics Data Booklet (2025); IB Physics subject reports and specimen papers (2023–2025). Last reviewed 2026-07-21.

Take this whole topic with you

Step-by-step worked examples — Wave model

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

Question type:

1Using the wave equation v = fλ
Getting startedDirect calculation• wave equation, AO2
Question
A wave on a string has a frequency of 5.0 Hz and a wavelength of 0.40 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 given values and evaluate.
  $v = 5.0 \times 0.40 = 2.0\,\text{m s}^{-1}$
Answer
v = 2.0 m s⁻¹.
Examiner tip
Mark scheme: (1) correct equation v = fλ; (2) v = 2.0 m s⁻¹ with unit. A one-step question, but the unit is worth a mark — a bare '2.0' is not fully credited.
2Period, frequency and speed together
Getting startedDirect calculation• period & frequency, AO2
Question
A water wave has a period of 0.25 s and a wavelength of 1.5 m. Calculate (a) its frequency and (b) its speed. (3 marks)
Step-by-step solution
1. Step 1
  Frequency is the reciprocal of the period: $f = 1/T$ .
  $f = \frac{1}{T} = \frac{1}{0.25} = 4.0\,\text{Hz}$
2. Step 2
  Now use the wave equation for the speed.
  $v = f\lambda = 4.0 \times 1.5 = 6.0\,\text{m s}^{-1}$
Answer
(a) f = 4.0 Hz; (b) v = 6.0 m s⁻¹.
Examiner tip
Mark scheme: (1) f = 1/T = 4.0 Hz; (2) v = fλ used; (3) v = 6.0 m s⁻¹ with unit. Converting the period to a frequency first is the key step — students who put T straight into v = fλ get a wrong answer.
3Classifying waves as transverse or longitudinal
Getting startedIdentify & classify• transverse vs longitudinal, AO1
Question
For each of the following, state whether it is transverse or longitudinal: (a) a sound wave in air, (b) visible light, (c) a wave on a stretched string, (d) a compression wave sent along a slinky spring. (4 marks)
Step-by-step solution
1. Step 1
  Ask for each: do the particles oscillate along the direction of travel (longitudinal) or perpendicular to it (transverse)?
2. Step 2
  (a) Sound = longitudinal (compressions and rarefactions). (b) Light = transverse (an EM wave). (c) Wave on a string = transverse (string moves up and down). (d) Slinky compression wave = longitudinal (coils move back and forth along the spring).
Answer
(a) longitudinal; (b) transverse; (c) transverse; (d) longitudinal.
Examiner tip
Mark scheme: 1 mark each. The reliable test is the direction of particle oscillation relative to the energy transfer. Sound and slinky-compression waves are the usual longitudinal examples; light and string waves are transverse.
4Reading wavelength and amplitude from a distance graph
Building confidenceGraph or diagram• graphs, AO2, AO3
Question
A displacement–distance graph of a wave shows crests 0.80 m apart, and each crest reaches 3.0 cm above the rest line. The wave travels at 24 m s⁻¹. Determine (a) the wavelength, (b) the amplitude and (c) the frequency of the wave. (4 marks)
Step-by-step solution
1. Step 1
  On a displacement–distance graph the horizontal gap between adjacent crests is the wavelength.
  $\lambda = 0.80\,\text{m}$
2. Step 2
  Amplitude is the maximum displacement from the rest line to a crest.
  $A = 3.0\,\text{cm} = 0.030\,\text{m}$
3. Step 3
  Rearrange the wave equation for frequency: $f = v/\lambda$ .
  $f = \frac{v}{\lambda} = \frac{24}{0.80} = 30\,\text{Hz}$
Answer
(a) λ = 0.80 m; (b) A = 3.0 cm = 0.030 m; (c) f = 30 Hz.
Examiner tip
Mark scheme: (1) λ = 0.80 m read from the distance axis; (2) A = 3.0 cm (from rest line, not peak-to-peak); (3) f = v/λ; (4) f = 30 Hz with unit. The distance axis means the crest spacing is the wavelength — this graph does NOT give the period.
5Reading period and frequency from a time graph
Building confidenceGraph or diagram• graphs, AO2, AO3
Question
A displacement–time graph for a single particle in a wave shows that successive crests occur 0.40 s apart. The wave has a speed of 15 m s⁻¹. Determine (a) the period, (b) the frequency and (c) the wavelength of the wave. (4 marks)
Step-by-step solution
1. Step 1
  On a displacement–time graph the horizontal gap between adjacent crests is the period (NOT the wavelength).
  $T = 0.40\,\text{s}$
2. Step 2
  Frequency is the reciprocal of the period.
  $f = \frac{1}{T} = \frac{1}{0.40} = 2.5\,\text{Hz}$
3. Step 3
  Now use the wave equation to find the wavelength: $\lambda = v/f$ .
  $\lambda = \frac{v}{f} = \frac{15}{2.5} = 6.0\,\text{m}$
Answer
(a) T = 0.40 s; (b) f = 2.5 Hz; (c) λ = 6.0 m.
Examiner tip
Mark scheme: (1) T = 0.40 s from the time axis; (2) f = 1/T = 2.5 Hz; (3) λ = v/f; (4) λ = 6.0 m with unit. The trap the examiner sets is to call 0.40 s a wavelength — on a time-axis graph it is the period.
6Frequency of an electromagnetic wave
Building confidenceDirect calculation• EM spectrum, AO2
Question
Red light has a wavelength of 700 nm in a vacuum. Taking the speed of light as c = 3.0×10⁸ m s⁻¹, calculate its frequency. (3 marks)
Step-by-step solution
1. Step 1
  Convert the wavelength to metres: $700\,\text{nm} = 700 \times 10^{-9}\,\text{m} = 7.0 \times 10^{-7}\,\text{m}$ .
  $\lambda = 7.0 \times 10^{-7}\,\text{m}$
2. Step 2
  For an EM wave in a vacuum $v = c$ , so rearrange $c = f\lambda$ for frequency.
  $f = \frac{c}{\lambda} = \frac{3.0 \times 10^{8}}{7.0 \times 10^{-7}}$
3. Step 3
  Evaluate.
  $f = 4.3 \times 10^{14}\,\text{Hz}$
Answer
f ≈ 4.3 × 10¹⁴ Hz.
Examiner tip
Mark scheme: (1) convert 700 nm to 7.0×10⁻⁷ m; (2) use c = fλ with c = 3.0×10⁸; (3) f = 4.3×10¹⁴ Hz. The nm → m conversion is where most marks are lost — a factor of 10⁹ error is very common.
7Combining both graphs to find the wave speed
StretchMulti-step problem• graphs, AO3, wave equation
Question
The same wave is shown on two graphs. Its displacement–distance graph shows adjacent crests 0.50 m apart. Its displacement–time graph shows adjacent crests 0.20 s apart. Determine (a) the frequency and (b) the speed of the wave, and explain which graph gives which quantity. (4 marks)
Step-by-step solution
1. Step 1
  Identify each quantity from the correct graph: the distance graph gives the wavelength; the time graph gives the period.
  $\lambda = 0.50\,\text{m}, \qquad T = 0.20\,\text{s}$
2. Step 2
  Frequency is the reciprocal of the period.
  $f = \frac{1}{T} = \frac{1}{0.20} = 5.0\,\text{Hz}$
3. Step 3
  Now combine to find the speed, using either $v = f\lambda$ or $v = \lambda/T$ .
  $v = \frac{\lambda}{T} = \frac{0.50}{0.20} = 2.5\,\text{m s}^{-1}$
Answer
(a) f = 5.0 Hz; (b) v = 2.5 m s⁻¹. The distance graph gives λ = 0.50 m; the time graph gives T = 0.20 s.
Examiner tip
Mark scheme: (1) λ from distance graph AND T from time graph correctly identified; (2) f = 1/T = 5.0 Hz; (3) v = λ/T; (4) v = 2.5 m s⁻¹ with unit. The whole point is not to mix the graphs up — reading 0.20 s as a wavelength or 0.50 m as a period wrecks the answer.
8Speed of sound from an echo
StretchWord problem• sound, AO2, AO3
Question
A student stands 170 m from a large cliff and claps once. She hears the echo 1.0 s later. (a) Calculate the speed of sound in air. (b) If the clap has a dominant frequency of 170 Hz, calculate the wavelength of the sound. (4 marks)
Step-by-step solution
1. Step 1
  The sound travels to the cliff and back, so the total distance is twice the separation.
  $d = 2 \times 170 = 340\,\text{m}$
2. Step 2
  Speed = total distance ÷ total time.
  $v = \frac{d}{t} = \frac{340}{1.0} = 340\,\text{m s}^{-1}$
3. Step 3
  Now use the wave equation, rearranged for wavelength: $\lambda = v/f$ .
  $\lambda = \frac{v}{f} = \frac{340}{170} = 2.0\,\text{m}$
Answer
(a) v = 340 m s⁻¹; (b) λ = 2.0 m.
Examiner tip
Mark scheme: (1) recognise the sound covers 2×170 = 340 m; (2) v = 340 m s⁻¹; (3) λ = v/f used; (4) λ = 2.0 m with unit. Forgetting the 'there and back' doubling is the classic echo error and halves the speed.

Model Answers — Wave model

High-scoring sample answers for wave model 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). State what is meant by a wave, and describe what a wave transfers.
Model answer
A wave is a disturbance that travels through space or a medium, transferring energy from one place to another without a net transfer of matter.

The particles of the medium simply oscillate about fixed positions — they do not travel along with the wave.
Why this scores
Why this scores 2/2. (1) a wave transfers energy; (2) without a net transfer of matter / particles only oscillate about fixed points. The phrase 'without a net transfer of matter' is the mark-earning idea — 'a wave moves through a medium' alone is not enough.
Question 2
3 marks
Q (3 marks). Distinguish between a transverse wave and a longitudinal wave, and give one example of each.
Model answer
In a transverse wave the particles oscillate perpendicular (at right angles) to the direction in which the energy is transferred. Example: light (or a wave on a string / a water surface wave).

In a longitudinal wave the particles oscillate parallel to the direction of energy transfer, producing compressions and rarefactions. Example: sound.
Why this scores
Why this scores 3/3. (1) transverse = oscillation perpendicular to energy transfer; (2) longitudinal = oscillation parallel, with compressions and rarefactions; (3) one correct example of each. The key wording is the direction of oscillation relative to the energy transfer.
Question 3
3 marks
Q (3 marks). A microwave oven produces microwaves of frequency 2.5 GHz. Taking the speed of the microwaves as 3.0×10⁸ m s⁻¹, calculate their wavelength.
Model answer
Convert the frequency to hertz: $2.5\,\text{GHz} = 2.5 \times 10^{9}\,\text{Hz}$ .

Microwaves are EM waves, so $v = c = 3.0\times10^{8}\,\text{m s}^{-1}$ . Rearrange the wave equation for wavelength:

$\lambda = \frac{v}{f} = \frac{3.0 \times 10^{8}}{2.5 \times 10^{9}}$

$\lambda = 0.12\,\text{m}\ (12\,\text{cm})$
Why this scores
Why this scores 3/3. (1) convert 2.5 GHz to 2.5×10⁹ Hz; (2) rearrange to λ = v/f; (3) λ = 0.12 m with unit. The GHz → Hz conversion (×10⁹) is the step examiners see fail most often.
Question 4
4 marks
Q (4 marks). (a) List the regions of the electromagnetic spectrum in order of increasing frequency. (b) State two properties that all electromagnetic waves have in common.
Model answer
(a) In order of increasing frequency (decreasing wavelength):

radio → microwave → infrared → visible → ultraviolet → X-ray → gamma.

(b) All electromagnetic waves:

are transverse waves, and

travel at the same speed $c = 3.0\times10^{8}\,\text{m s}^{-1}$ in a vacuum.

(They also all transfer energy without a medium.)
Why this scores
Why this scores 4/4. (a) 2 marks for the correct full order (1 mark if one or two regions are misplaced); (b) 1 mark for 'transverse' and 1 mark for 'all travel at c in a vacuum'. A frequent error is to think higher-frequency waves travel faster — they do not.
Question 5
5 marks
Q (5 marks). A wave is represented by two graphs. On the displacement–distance graph, adjacent crests are 2.0 m apart. On the displacement–time graph for one particle, adjacent crests are 0.50 s apart. (a) State which quantity each graph gives and its value. (b) Calculate the frequency. (c) Calculate the speed of the wave.
Model answer
(a) The displacement–distance graph gives the wavelength: $\lambda = 2.0\,\text{m}$ . The displacement–time graph gives the period: $T = 0.50\,\text{s}$ .

(b) Frequency is the reciprocal of the period: $f = \frac{1}{T} = \frac{1}{0.50} = 2.0\,\text{Hz}$

(c) Using the wave equation: $v = f\lambda = 2.0 \times 2.0 = 4.0\,\text{m s}^{-1}$
Why this scores
Why this scores 5/5. (a) λ = 2.0 m from the distance graph (1) and T = 0.50 s from the time graph (1); (b) f = 1/T = 2.0 Hz (1); (c) v = fλ used (1), v = 4.0 m s⁻¹ (1). The assessed skill is knowing that the distance graph gives λ and the time graph gives T — never the other way round.
Question 6
4 marks
Q (4 marks). Explain why sound cannot travel through a vacuum, but light from the Sun reaches the Earth through the vacuum of space. Refer to the type of wave in each case.
Model answer
Sound is a longitudinal mechanical wave: it travels as compressions and rarefactions of the particles of a medium (e.g. air). A vacuum contains no particles, so there is nothing to compress and pass the disturbance on — therefore sound cannot travel through it.

Light is an electromagnetic wave: it is an oscillation of electric and magnetic fields and does not need a medium of particles to travel. It can therefore cross the vacuum of space, which is how sunlight reaches the Earth.
Why this scores
Why this scores 4/4. (1) sound is a mechanical wave needing particles of a medium; (2) a vacuum has no particles so sound cannot pass; (3) light is an EM wave; (4) EM waves need no medium and cross a vacuum. Answers must contrast mechanical (needs medium) with electromagnetic (does not).
Question 7
6 marks
Q (6 marks). A transverse water wave travels at 0.60 m s⁻¹. Its displacement–distance graph shows crests 0.30 m apart, each reaching 0.020 m above the rest line. Determine (a) the amplitude, (b) the wavelength, (c) the frequency and (d) the period of the wave, and (e) state how many complete waves pass a fixed point in 5.0 s.
Model answer
(a) Amplitude = maximum displacement from the rest line: $A = 0.020\,\text{m}$ .

(b) On a displacement–distance graph the crest spacing is the wavelength: $\lambda = 0.30\,\text{m}$ .

(c) Rearranging the wave equation for frequency: $f = \frac{v}{\lambda} = \frac{0.60}{0.30} = 2.0\,\text{Hz}$

(d) Period is the reciprocal of the frequency: $T = \frac{1}{f} = \frac{1}{2.0} = 0.50\,\text{s}$

(e) Number of waves in 5.0 s = $f \times t = 2.0 \times 5.0 = 10$ complete waves.
Why this scores
Why this scores 6/6. (1) A = 0.020 m from the rest line; (2) λ = 0.30 m from the distance graph; (3) f = v/λ = 2.0 Hz; (4) T = 1/f = 0.50 s; (5) correct method for part (e); (6) 10 waves. Part (e) tests that frequency literally means 'waves per second' — 2.0 Hz × 5.0 s = 10.
Question 8
6 marks
Q (6 marks). An X-ray used in medical imaging has a wavelength of 1.0×10⁻¹⁰ m. A radio wave used for broadcasting has a frequency of 9.0×10⁷ Hz. Both travel in a vacuum where c = 3.0×10⁸ m s⁻¹. (a) Calculate the frequency of the X-ray. (b) Calculate the wavelength of the radio wave. (c) State which of the two waves travels faster, and justify your answer.
Model answer
(a) For the X-ray, rearrange $c = f\lambda$ for frequency: $f = \frac{c}{\lambda} = \frac{3.0\times10^{8}}{1.0\times10^{-10}} = 3.0\times10^{18}\,\text{Hz}$

(b) For the radio wave, rearrange $c = f\lambda$ for wavelength: $\lambda = \frac{c}{f} = \frac{3.0\times10^{8}}{9.0\times10^{7}} = 3.3\,\text{m}$

(c) Neither is faster — they travel at the same speed. Both are electromagnetic waves, and all EM waves travel at $c = 3.0\times10^{8}\,\text{m s}^{-1}$ in a vacuum. They differ only in frequency and wavelength, not in speed.
Why this scores
Why this scores 6/6. (a) f = c/λ = 3.0×10¹⁸ Hz with unit (2); (b) λ = c/f = 3.3 m with unit (2); (c) both travel at the same speed c because all EM waves do (2). The examiner's trap in part (c) is to expect a 'faster' answer — the mark is for recognising the speeds are equal.
Question 9
10 marks
Q (10 marks — extended response). A loudspeaker emits a sound of frequency 256 Hz into air, where the speed of sound is 340 m s⁻¹. (a) State the type of wave that sound is and describe how the air moves as the sound passes. (b) Calculate the wavelength of the sound. (c) The same information (frequency 256 Hz) is sent as a radio wave to a distant receiver. Taking c = 3.0×10⁸ m s⁻¹, calculate the wavelength of this radio wave and comment on how it compares with the sound wavelength. (d) Explain why the radio signal reaches the receiver but the sound would not, if the receiver were in space. (e) A displacement–time graph of a single air particle for the sound shows an amplitude of 1.2×10⁻⁵ m. State what the amplitude represents and what a larger amplitude would mean for the sound heard.
Model answer
(a) Sound is a longitudinal mechanical wave. As it passes, the air particles oscillate back and forth about fixed positions, parallel to the direction the wave travels, creating regions of compression (particles close together, higher pressure) and rarefaction (particles spread out, lower pressure). The particles do not travel with the wave — only the energy does.

(b) Using the wave equation $v = f\lambda$ , rearranged for wavelength: $\lambda_{\text{sound}} = \frac{v}{f} = \frac{340}{256} = 1.33\,\text{m}$

(c) For the radio wave, $v = c = 3.0\times10^{8}\,\text{m s}^{-1}$ : $\lambda_{\text{radio}} = \frac{c}{f} = \frac{3.0\times10^{8}}{256} = 1.17\times10^{6}\,\text{m}$ This is about $1.2\times10^{6}\,\text{m}$ — roughly a million times longer than the sound wavelength. Although both waves have the same frequency, the radio wave travels almost a million times faster, so from $\lambda = v/f$ its wavelength is almost a million times greater.

(d) A radio wave is an electromagnetic wave, which is an oscillation of electric and magnetic fields and needs no medium — so it can travel through the vacuum of space to the receiver. Sound is a mechanical wave that needs a medium of particles to carry the compressions and rarefactions; space is a vacuum with no particles, so the sound has nothing to travel through and cannot reach the receiver.

(e) The amplitude is the maximum displacement of an air particle from its rest position. A larger amplitude means each particle oscillates further, the wave carries more energy, and the sound is heard as louder.
Why this scores
Why this scores 10/10. (a) longitudinal mechanical wave + particles oscillate parallel to travel forming compressions/rarefactions, energy not matter transferred (2); (b) λ = v/f = 1.33 m (2); (c) λ_radio = c/f ≈ 1.2×10⁶ m, correctly noting same frequency but ~10⁶× longer because the speed is ~10⁶× greater (2); (d) radio = EM wave needs no medium so crosses the vacuum, sound = mechanical wave needs particles which a vacuum lacks (2); (e) amplitude = max displacement from rest, larger amplitude → more energy → louder (2). The grade-9 discriminators are (i) holding the frequency fixed while the speed and wavelength both change between the two media, and (ii) linking amplitude to energy and loudness rather than to pitch.

Key Formulae — Wave model

The formulae you need to memorise for wave model 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
The central relationship for every wave — use whenever two of speed, frequency and wavelength are known and the third is wanted.
Example
f = 5.0 Hz, λ = 0.40 m → v = 5.0 × 0.40 = 2.0 m s⁻¹.
Period–frequency relation
$T = \frac{1}{f}$
$T$
period — time for one oscillation (s)
$f$
frequency — oscillations per second (Hz)
When to use
To convert between period and frequency; equivalently f = 1/T. Needed before reading a time graph's period into the wave equation.
Example
T = 0.25 s → f = 1/0.25 = 4.0 Hz.
Wave speed from wavelength and period
$v = \frac{\lambda}{T}$
$v$
wave speed (m s⁻¹)
$\lambda$
wavelength (m)
$T$
period (s)
When to use
When you have the wavelength (from a distance graph) and the period (from a time graph) — combine them directly without finding f first.
Example
λ = 0.50 m, T = 0.20 s → v = 0.50/0.20 = 2.5 m s⁻¹.
Speed of electromagnetic waves in a vacuum
$c = f\lambda = 3.0\times10^{8}\,\text{m s}^{-1}$
$c$
speed of light in a vacuum (m s⁻¹)
$f$
frequency of the EM wave (Hz)
$\lambda$
wavelength of the EM wave (m)
When to use
For any electromagnetic wave (radio to gamma) travelling in a vacuum or air — the speed is always c, so use c = fλ to find the missing frequency or wavelength.
Example
λ = 7.0×10⁻⁷ m → f = c/λ = 3.0×10⁸ / 7.0×10⁻⁷ = 4.3×10¹⁴ Hz.

Key Definitions and Keywords — Wave model

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

Wave
Examiner keyword
A disturbance that transfers energy from one place to another without a net transfer of matter; the particles of the medium oscillate about fixed positions.
Transverse wave
Examiner keyword
A wave in which the particles oscillate perpendicular (at right angles) to the direction of energy transfer. Examples: light and all EM waves, waves on a string, water surface waves.
Longitudinal wave
Examiner keyword
A wave in which the particles oscillate parallel to the direction of energy transfer, producing compressions and rarefactions. Example: sound.
Wavelength (λ)
Examiner keyword
The distance between two adjacent points on a wave that are in phase (e.g. crest to next crest). SI unit: metre (m).
Frequency (f)
Examiner keyword
The number of complete oscillations (waves) passing a point per second. SI unit: hertz (Hz), equal to s⁻¹.
Period (T)
Examiner keyword
The time taken for one complete oscillation of the wave. Related to frequency by T = 1/f. SI unit: second (s).
Amplitude (A)
Examiner keyword
The maximum displacement of a particle from its rest (equilibrium) position. Measured from the middle line to a crest, not peak-to-peak. SI unit: metre (m).
Wave speed (v)
Examiner keyword
The speed at which the wave (its energy) travels through the medium. Given by v = fλ. SI unit: m s⁻¹.
In phase
Examiner keyword
Two points on a wave are in phase when they are moving in exactly the same way at the same time (e.g. two crests). Adjacent in-phase points are one wavelength apart.
Compression
Examiner keyword
A region of a longitudinal wave where the particles are pushed close together, giving higher pressure (e.g. in a sound wave).
Rarefaction
Examiner keyword
A region of a longitudinal wave where the particles are spread apart, giving lower pressure (e.g. in a sound wave).
Wavefront
Examiner keyword
A line or surface joining all points on a wave that are in phase (e.g. along the top of a crest). Adjacent wavefronts are one wavelength apart.
Ray
Examiner keyword
A line drawn perpendicular to the wavefronts showing the direction in which the wave (energy) travels.
Electromagnetic spectrum
Examiner keyword
The full family of electromagnetic waves, all transverse and all travelling at c = 3.0×10⁸ m s⁻¹ in a vacuum, ordered by increasing frequency: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma.
Mechanical wave
Examiner keyword
A wave that requires a medium of particles to travel through (e.g. sound, water and seismic waves). It cannot pass through a vacuum.
Electromagnetic wave
Examiner keyword
A transverse wave consisting of oscillating electric and magnetic fields that needs no medium and travels at c = 3.0×10⁸ m s⁻¹ in a vacuum.

Common Mistakes and Misconceptions — Wave model

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

✕Reading the wavelength off a displacement–time graph
IB Physics Theme C subject reports
Why it happens
The two graphs look identical (both are sine curves), so students grab the crest-to-crest gap without checking the horizontal axis.
How to avoid it
Check the horizontal axis first. On a time axis the crest-to-crest gap is the period T (then f = 1/T); only a distance axis gives the wavelength λ.
✕Confusing frequency with period
IB Physics Theme C subject reports
Why it happens
Both describe how often the wave oscillates, and students forget which is time and which is a rate.
How to avoid it
Period T is a time (seconds) for one wave; frequency f is a rate (waves per second, Hz). They are reciprocals: T = 1/f. Convert a period to a frequency before using v = fλ.
✕Thinking the particles travel along with the wave
IB Physics Theme C subject reports
Why it happens
The wave visibly moves forward, so students assume the medium moves with it.
How to avoid it
State clearly that a wave transfers energy, not matter. Each particle only oscillates about a fixed rest position; it does not move along with the wave (think of a bobbing duck).
✕Mixing up transverse and longitudinal examples
IB Physics Theme C subject reports
Why it happens
Students memorise the definitions but not which everyday waves belong to each class — sound is especially often mislabelled as transverse.
How to avoid it
Anchor the examples: sound = longitudinal; light and all EM waves = transverse; waves on a string and water surface waves are transverse. Test each by asking which way the particles oscillate relative to the energy transfer.
✕Thinking different EM waves travel at different speeds
IB Physics Theme C subject reports
Why it happens
Because gamma rays are 'higher energy', students assume they are also faster than radio waves.
How to avoid it
Remember that all EM waves travel at the same speed c = 3.0×10⁸ m s⁻¹ in a vacuum. They differ only in frequency and wavelength (linked by c = fλ), never in speed.
✕Unit and prefix slips (nm, MHz, GHz) in wave calculations
IB Physics Theme C subject reports
Why it happens
Wave problems mix very large and very small numbers, and students forget to convert prefixes to SI before substituting.
How to avoid it
Always convert to base SI first: 1 nm = 10⁻⁹ m, 1 MHz = 10⁶ Hz, 1 GHz = 10⁹ Hz. Substitute the converted values into v = fλ and quote the final answer with its unit.

C.2 Wave model — IB Physics SL Study Notes (Theme C: Wave Behaviour)

Feature

Transverse wave

Longitudinal wave

Oscillation direction

Perpendicular to energy transfer

Parallel to energy transfer

Shape

crests and troughs

compressions and rarefactions

Examples

light & all EM waves, waves on a string, water surface waves

sound, a compression wave on a slinky, seismic P-waves

Needs a medium?

EM waves do not; string/water do

Yes — always a mechanical wave

Quantity

Symbol

Meaning

SI unit

Wavelength

\lambda

Distance between two adjacent points in phase (e.g. crest to next crest)

Frequency

f

Number of complete oscillations (waves) passing a point per second

Hz (s⁻¹)

Period

T

Time for one complete oscillation

Amplitude

A

Maximum displacement of a particle from its rest position

Wave speed

v

Speed at which the wave (energy) travels

m s⁻¹

Sound is the most important example of a longitudinal mechanical wave. A vibrating source (a loudspeaker cone, a guitar string, your vocal cords) pushes on the air:

as it moves forward it squashes the air particles together → a compression (high pressure);
as it moves back it lets them spread apart → a rarefaction (low pressure).

Wave model

Detailed Study Notes

At a glance

What you’ll learn

What a wave is: energy on the move, matter staying put

The five wave quantities you must know

Deriving the wave equation v = fλ from first principles

The two graphs: displacement–distance vs displacement–time

Wavefronts, rays and the electromagnetic spectrum

Sound: a longitudinal mechanical wave

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Using the wave equation v = fλ

2Period, frequency and speed together

3Classifying waves as transverse or longitudinal

4Reading wavelength and amplitude from a distance graph

5Reading period and frequency from a time graph

6Frequency of an electromagnetic wave

7Combining both graphs to find the wave speed

8Speed of sound from an echo

The wave equation

Period–frequency relation

Wave speed from wavelength and period

Speed of electromagnetic waves in a vacuum

Wave

Transverse wave

Longitudinal wave

Wavelength (λ)

Frequency (f)

Period (T)

Amplitude (A)

Wave speed (v)

In phase

Compression

Rarefaction

Wavefront

Ray

Electromagnetic spectrum

Mechanical wave

Electromagnetic wave

✕Reading the wavelength off a displacement–time graph

✕Confusing frequency with period

✕Thinking the particles travel along with the wave

✕Mixing up transverse and longitudinal examples

✕Thinking different EM waves travel at different speeds

✕Unit and prefix slips (nm, MHz, GHz) in wave calculations

Wave model

Detailed Study Notes

At a glance

What you’ll learn

What a wave is: energy on the move, matter staying put

The five wave quantities you must know

Deriving the wave equation v = fλ from first principles

The two graphs: displacement–distance vs displacement–time

Wavefronts, rays and the electromagnetic spectrum

Sound: a longitudinal mechanical wave

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Using the wave equation v = fλ

2Period, frequency and speed together

3Classifying waves as transverse or longitudinal

4Reading wavelength and amplitude from a distance graph

5Reading period and frequency from a time graph

6Frequency of an electromagnetic wave

7Combining both graphs to find the wave speed

8Speed of sound from an echo