AQA A-level Physics 7408 · Paper 1 · Spec 3.3.1

Progressive and Stationary Waves

See the particles of a wave oscillate, build a stationary wave from two travelling waves, and run required practical 1 on a virtual string. Free simulations, exam questions and a printable booklet.

On this page: ● node (N), ○ antinode (A) · — wave travelling right · - - wave travelling left

What is moving in a wave?

The wave moves along, but each particle only oscillates about its own rest position. Switch between a transverse and a longitudinal wave, and compare particles P and Q. The graph on the right follows P and Q over time.

Wave speed c = fλ
Period T = 1f
Phase difference between P and Q
P and Q
Displacement–time graph for P and Q
Each curve follows one particle as time passes: peak to peak is the period T. Q reaches each crest a time Δt after P, so the phase difference is 2πΔtT. This graph has time on its axis, not distance: read the axis label first.

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More particles per metre means more mass per metre, μ, so with the same tension the wave is slower: c = √Tμ. The wavelength then follows from λ = cf. Displacements are drawn larger than scale so you can see them. The wave travels to the right. In the longitudinal view a displacement to the right counts as positive on the graph.

Polarisation: evidence that a wave is transverse

A polarising filter only lets through oscillations in one plane. Turn the second filter, the receiving aerial or the slotted screen, and see which waves are affected.

Transmitter
Two Polaroid filters (the arrows show the oscillations, looking along the beam)

Brightness and signal are shown in words and a bar, not numbers: AQA does not ask you to calculate how much light gets through at an angle.

Building a stationary wave

A wave travelling right meets its reflection travelling left. They have the same frequency and amplitude, and they superpose. Step through one period in eighths and watch the nodes stay still.

Top: the two travelling waves · Bottom: their sum (string fixed at both ends, l = 1.20 m)
Time
Wavelength λ = 2ln
Node spacing λ2
X and Y

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String lab: required practical 1

A vibration generator drives a string over a pulley, with masses hanging from the end. Turn the signal generator until the string resonates, then record the reading. Change one variable at a time and plot a straight-line graph.

The apparatus
Tension T = mg
Mass per unit length μ
Generator f
Resonance?
Your results
Recorded resonances
l / mT / Nμ / kg m−1nf / Hzf1 / Hz
Record at least three resonances, changing only one variable, to see a straight line.

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g = 9.81 N kg−1. The generator end of the string is very nearly a node, because the vibration generator moves only a tiny amount. Safety in the real practical: keep your feet clear of the hanging masses and wear eye protection in case the wire snaps.

Key equations for progressive and stationary waves

Wave speed
AQA formula sheet
c = fλ
Speed in m s−1, frequency in Hz, wavelength in m
Frequency and period
AQA formula sheet
f = 1T
Convert ms to s before you divide
First harmonic on a string
AQA formula sheet
f = 12l√Tμ
T is the tension in N; μ is the mass per unit length in kg m−1
Phase difference
Useful relationship: learn it
Δφ = 2πdλ
Progressive waves only. In radians; 2π rad = 360° = one cycle
Harmonics on a string
Useful relationship: derive it
λ = 2ln
and fn = nf1. First harmonic: λ = 2l
Node spacing
Useful relationship: learn it
N to N = λ2
Node to the next antinode is λ4
Mass per unit length
Useful relationship
μ = mL = ρA
Not density: kg m−1, not kg m−3

Quick catch

Three rounds of five multiple-choice questions, each round harder than the last. Pick A, B, C or D and the reasons appear straight away. Get 4 of 5 right to unlock the next round.

0 of 5 right

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Where it comes from

Three ideas behind stationary waves, built up step by step. AQA expects the graphical explanation and the harmonics; the third tab shows the algebra behind them.

1
Two waves, opposite directions
A wave travels along a string and reflects at the fixed end. The reflected wave has the same frequency, wavelength and (nearly) the same amplitude, but travels the other way.
2
Superposition at every point
Where the two waves meet, the displacement is the sum of their displacements. Draw both waves at one instant and add them point by point.
y = y1 + y2
3
Some points always cancel
Step on by T8 each time. The waves move λ8 in opposite directions, so at certain points they always arrive in antiphase and cancel: these are the nodes. At T4 the whole string is flat for an instant.
4
Some points always reinforce
Halfway between the nodes the waves always arrive in phase, so the string swings with twice the amplitude of either wave: these are the antinodes.
N to N = λ2
Snapshots through half a period
t = 0t =T8t =T4t =3T8t =T2wave to the rightwave to the leftsum
The nodes (dots) never move. Between t = 0 and t = T2 the sum goes from one extreme, through flat, to the other.
1
Fixed ends are nodes
The string cannot move where it is clamped, so there must be a node at each end.
2
Fit whole loops
Each loop is half a wavelength. A string of length l can hold n loops, where n = 1, 2, 3 …
l = n λ2
3
The allowed wavelengths
Rearrange. The first harmonic (n = 1) has the longest wavelength, twice the length of the string.
λ = 2ln
4
The frequencies
The wave speed c = √Tμ depends only on the tension and the string, so it is the same for every harmonic. With f = cλ:
fn = n2l√Tμ = nf1
The first three harmonics
n = 1λ = 2ln = 2λ = ln = 3λ =2l3
Example: a string of length 0.65 m, tension 70 N, μ = 1.2 × 10−3 kg m−1 has f1 = 11.30 × √701.2 × 10−3 = 186 Hz, so its third harmonic is 557 Hz.
Not examined by AQA. This uses a trigonometry identity from A-level Maths to show why the nodes stay still.
1
Write the two waves
With ω = 2πf and k = 2πλ, a wave travelling right and its inverted reflection travelling left are:
y1 = A sin(ωt − kx)   y2 = −A sin(ωt + kx)
2
Add them
Use sin P − sin Q = 2 cosP + Q2 sinP − Q2.
y = −2A sin(kx) cos(ωt)
3
Read it
Every point oscillates with the same cos(ωt), so all points in one loop are in phase. The amplitude, 2A |sin kx|, depends on position: zero where sin kx = 0.
nodes at x = 0, λ2, λ, …
4
Antiphase across a node
sin kx changes sign at each node, so points in neighbouring loops are always in antiphase. Nothing travels: there is no (ωt − kx) term left.
Amplitude along the string
amplitude 2A|sin kx|xdashed: the string at one instant
Amplitude 2A|sin kx|: zero at the nodes, 2A at the antinodes. The sign tells you which way each loop is moving.

Test yourself

35 exam-style questions in three levels: Core, Exam-style and Stretch for A*. Each level shows a few at a time. Press Try another question for a different one. Type numbers without units. For the explain parts, write your answer, then mark it against the mark scheme.

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Download the booklet

Notes, worked examples, equation cards, 17 quick checks and 32 exam-style questions on progressive and stationary waves, including required practical 1, to print and write on. The answers are at the back.

Progressive and Stationary Waves: notes and practice booklet

Print it, write in the spaces, then check your answers at the back.

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  • 14 October 2026
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It opens in a new tab. Use the download or print button in the PDF viewer to keep a copy. Print it double-sided and write in the spaces. In the PDF, each question links to its answer and back again.

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Progressive and stationary waves: common questions

Why is the first-harmonic wavelength twice the length of the string?

Both ends are clamped, so both must be nodes. The simplest pattern between two nodes is a single loop, and one loop is half a wavelength.

l = λ2  so  λ = 2l

A guitar string 0.65 m long therefore has a first-harmonic wavelength of 1.30 m. The next harmonics have two and three loops, with λ = l and 2l3.

Exam trap. Using λ = l for the first harmonic halves every wave speed and frequency you calculate.

Do stationary waves carry energy?

A stationary wave stores energy, but it transfers no net energy along the string. It is made of two equal waves travelling in opposite directions, so their energy flows cancel.

Inside each loop, energy moves back and forth between kinetic energy (when the string is flat and moving fastest) and potential energy in the stretched string (at the extremes). Nodes never move, so energy cannot cross them.

Exam trap. In a comparison table, write “no net energy transfer” for the stationary wave, not just “no energy”.

Why can’t sound be polarised?

Polarising a wave means restricting its oscillations to one plane. That needs oscillations at right angles to the direction of travel, which only transverse waves have.

Sound is longitudinal: the air oscillates back and forth along the direction the sound travels. There is no sideways oscillation to restrict, so a slotted screen or filter changes nothing however you turn it. Try the third mode of the polarisation simulation.

Exam trap. “Only transverse waves can be polarised” earns the mark; “sound is a different type of wave” does not.

Why do Polaroid sunglasses cut glare?

Light reflected from water, wet roads and car bonnets is partly plane polarised, with its oscillations mostly horizontal. The sunglasses contain Polaroid with its transmission axis vertical.

The vertical Polaroid blocks much of the horizontally polarised glare, while still letting through about half of the unpolarised light from everything else. Tilt your head 90° with the sunglasses on and the glare comes back.

Exam trap. Say which direction the reflected light is polarised in and which way the filter is aligned: “at right angles” is the key idea.