OCR A Level Physics A: Stationary waves, harmonics and resonance tubes (H556)

This guide covers section 4.4.4 (Stationary waves), statements (a) to (g), of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. H556 is the full A level. The ideas on this page build on progressive waves and wavelength, which you should already know. Superposition and the double slit are taught elsewhere. Course pages: OCR A Level Physics hub | checklist | free diagnostics.

Companion pages: revision notes and practice questions.

What this unit covers

Statement	What you must be able to do
4.4.4(a)	Describe stationary (standing) waves using microwaves, stretched strings and air columns
4.4.4(b)	Interpret graphs that represent a stationary wave
4.4.4(c)	Compare stationary and progressive waves
4.4.4(d)	Locate and describe nodes and antinodes
4.4.4(e)(i)	Sketch and use the patterns for a stretched string and for air columns in closed and open tubes
4.4.4(e)(ii)	Describe the technique for finding the speed of sound in air with a resonance tube (linked to PAG5)
4.4.4(f)	Use the result that adjacent nodes (or antinodes) are lambda/2 apart
4.4.4(g)	Name the fundamental mode (1st harmonic) and the harmonics

How a stationary wave forms (4.4.4(a))

Superposing two progressive waves gives a stationary wave when they share a frequency and speed, have similar amplitudes, and move through the same region in opposite directions. In practice the second wave is the reflection of the first. The result is a pattern that stays in place. Energy is not carried along it.

Microwaves. A microwave transmitter faces a flat metal plate. The reflected wave returns along the path of the incident wave. A small receiver (a probe) moved along the line between transmitter and plate gives a reading that rises and falls. Maximum readings are antinodes. Minimum readings, close to zero, are nodes.

Stretched string. One end of a string is attached to a vibration generator driven by a signal generator. The other end passes over a pulley and carries a hanging mass. Waves travel from the generator, reflect at the fixed end and superpose with the oncoming waves. At particular frequencies, large steady loops appear.

Air columns. A loudspeaker (or a vibrating tuning fork) sends sound into a tube. The wave reflects from the far end and superposes with the incoming wave. At resonance the sound is noticeably loud.

Graphs of a stationary wave (4.4.4(b))

A graph of displacement against position, for one instant, looks like a sine curve. Draw it again a quarter of a period later and every particle is at zero displacement, so the line is flat. After half a period you get the mirror image of the first graph. Between these times the curve rises and falls inside a fixed outline, the envelope. The envelope is the pair of curves showing the maximum displacement at each position.

Key facts from the graph:

• the amplitude depends on position. It is zero at a node and greatest at an antinode
• the largest amplitude is the sum of the amplitudes of the two progressive waves
• all particles between two adjacent nodes move in phase with each other
• particles in adjacent loops (on either side of a node) are in antiphase
• all particles pass through their equilibrium position at the same instant

A graph of displacement against time, for one particle, is a sine curve with the same period at every position (except nodes, where it is a flat line). Only its amplitude differs.

Worked example 1: reading a graph

A stationary wave on a string has an antinode amplitude of 6.0 mm. Nodes are at 0, 20 cm, 40 cm and 60 cm.

1. Each progressive wave has half the maximum amplitude: 6.0 / 2 = 3.0 mm.
2. Adjacent nodes are 20 cm apart, so lambda/2 = 20 cm and lambda = 40 cm.
3. The points at 10 cm and 30 cm are in adjacent loops, so they move in antiphase. The points at 5 cm and 15 cm are in the same loop, so they move in phase, though with different amplitudes.

Stationary and progressive waves compared (4.4.4(c))

Feature	Progressive wave	Stationary wave
Energy	Transferred along the wave	Not transferred along the wave; stored in the oscillations
Amplitude	Same for every particle	Varies from zero (node) to maximum (antinode)
Phase	Changes steadily along the wave, one cycle per wavelength	Same within a loop; changes by 180 degrees across a node
Wave profile	Moves in the direction of travel	Stays fixed in position
Frequency	Every particle oscillates at the wave frequency	Every particle (except at a node) oscillates at the same frequency
Wavelength	Distance between neighbouring points in phase	Twice the distance between adjacent nodes

Similarities to quote: both can be produced by the same source, and both have a wavelength and a frequency linked by the same speed of the progressive waves, v = f lambda.

Nodes and antinodes (4.4.4(d)) and the lambda/2 rule (4.4.4(f))

A node is a point of zero amplitude: the two waves are always in antiphase there and cancel. An antinode is a point of maximum amplitude: the two waves are always in phase and add.

Adjacent nodes are separated by lambda/2, where lambda is the wavelength of the progressive waves. The same is true for adjacent antinodes. A node and the next antinode are lambda/4 apart.

Worked example 2: microwaves

A probe moved between a transmitter and a metal plate finds six successive minima. The first and sixth are 7.9 cm apart. The speed of electromagnetic waves is 3.00 x 10^8 m/s.

1. Six minima make 5 gaps, each lambda/2. So 5 x lambda/2 = 7.9 cm.
2. lambda = 2 x 7.9 / 5 = 3.16 cm = 3.16 x 10^-2 m.
3. f = v / lambda = 3.00 x 10^8 / 3.16 x 10^-2 = 9.5 x 10^9 Hz.

Counting minima instead of gaps is a very common slip. Measuring across many gaps also reduces the percentage uncertainty in lambda.

Patterns and harmonics (4.4.4(e)(i) and (g))

The fundamental mode of vibration (1st harmonic) is the pattern with the lowest frequency. Higher frequencies that also give stationary waves are the 2nd, 3rd and further harmonics. Resonance only occurs for these particular frequencies, because the pattern must fit the boundary conditions.

String fixed at both ends

Both ends are nodes. The length L holds a whole number of half-wavelengths.

Harmonic	Loops	Length	Wavelength	Frequency
1st	1	L = lambda/2	2L	f1 = v / 2L
2nd	2	L = lambda	L	2 f1
3rd	3	L = 3 lambda/2	2L/3	3 f1

In general the nth harmonic has n loops, wavelength 2L/n and frequency n f1.

Worked example 3: a string

A string of length 0.64 m is fixed at both ends and vibrates at its fundamental frequency of 110 Hz.

1. lambda = 2L = 2 x 0.64 = 1.28 m.
2. Wave speed v = f lambda = 110 x 1.28 = 141 m/s (140.8 to 3 significant figures).
3. The 2nd and 3rd harmonics are 220 Hz and 330 Hz.
4. In the 3rd harmonic, lambda = 2L/3 = 0.427 m, so adjacent nodes are 0.213 m apart. There are four nodes in all (two at the ends, two inside).

Tube closed at one end

The closed end is a node (air cannot move there). The open end is close to an antinode. The length holds an odd number of quarter-wavelengths.

Harmonic	Pattern	Length	Frequency
1st	node to antinode	L = lambda/4	f1 = v / 4L
3rd	node, antinode, node, antinode	L = 3 lambda/4	3 f1
5th		L = 5 lambda/4	5 f1

Even harmonics are missing. The frequencies are f1, 3 f1, 5 f1 and so on.

Tube open at both ends

Both ends are antinodes. The length holds a whole number of half-wavelengths, so the pattern has the same frequencies as the string: f1 = v / 2L, then 2 f1, 3 f1.

Worked example 4: air columns

A tube is 0.375 m long. Take the speed of sound as 340 m/s.

• Closed at one end: f1 = v / 4L = 340 / (4 x 0.375) = 227 Hz. The next resonances are 3 x 227 = 680 Hz and 5 x 227 = 1130 Hz.
• Open at both ends: f1 = v / 2L = 340 / 0.75 = 453 Hz, with the 2nd harmonic at 907 Hz.

The same tube therefore sounds an octave higher when it is open, because its fundamental frequency doubles.

Speed of sound with a resonance tube (4.4.4(e)(ii), PAG5)

Apparatus. A tall transparent tube stands in a tall cylinder of water. Raising or lowering the tube changes the length of the air column above the water surface. A tuning fork of known frequency f is struck and held over the open end.

Method.

1. Start with a short air column and lower the tube slowly while the fork sounds.
2. Find the length L1 at which the sound is loudest. Repeat a few times and take the mean.
3. Continue lowering until the sound is loud again and record L2.
4. Repeat with forks of different frequencies.

The water surface is the closed end (a node). The first resonance is the fundamental of a closed tube, and the second is the 3rd harmonic. The extra length between them is half a wavelength, so

lambda = 2 (L2 - L1) and v = f lambda = 2 f (L2 - L1).

Using the difference removes the need to know exactly where the antinode forms just above the open end, because that offset is the same in both readings. Using L1 alone gives lambda = 4 L1, which is slightly too small.

Worked example 5: finding v

A 512 Hz fork gives resonances at L1 = 15.9 cm and L2 = 48.6 cm.

1. L2 - L1 = 32.7 cm = 0.327 m, which equals lambda/2.
2. lambda = 0.654 m.
3. v = 512 x 0.654 = 335 m/s.
4. Check: lambda/4 = 0.1635 m, which is 0.0045 m more than L1. The antinode sits a few millimetres above the tube mouth.

Improving reliability. Listen for a clear maximum, approach each resonance from above and below, use several forks and take the mean v. A graph can also help (worked example 6).

Worked example 6: a graph of f against 1/L

For a string of fixed tension, the fundamental satisfies f = v / 2L. So a graph of f against 1/L is a straight line that passes through the origin, with gradient v / 2.

L / m	0.40	0.50	0.60	0.80	1.00
f / Hz	120	96	80	60	48
1/L / m^-1	2.50	2.00	1.67	1.25	1.00

Gradient = (120 - 48) / (2.50 - 1.00) = 48 m/s, so v = 2 x 48 = 96 m/s.

Common errors

• Counting nodes instead of gaps when finding lambda from the separation of several nodes.
• Saying a node is “where the wave stops”. It is where the displacement is always zero because the waves cancel.
• Writing that L = lambda for the fundamental of a string. It is lambda/2.
• Giving even harmonics for a closed tube.
• Saying particles in a stationary wave all have the same amplitude.
• Using the tube length L1 alone, or forgetting that the second resonance is the 3rd harmonic, not the 2nd.

Test yourself with the practice questions, then check your level with the free diagnostics.

Official syllabus

OCR Advanced GCE in Physics A (H556), Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Section 4.4.4, Stationary waves.