Standing Waves

When two progressive waves of the same frequency and amplitude travel in opposite directions through the same medium, they superpose to produce a very distinctive pattern called a standing wave (or stationary wave). Unlike progressive waves, standing waves do not transfer energy — they store it. Understanding standing waves is essential for explaining musical instruments, resonance, and for experimentally measuring the speed of sound.

Formation of Standing Waves

A standing wave forms when a wave reflects from a boundary and the reflected wave meets the incoming wave. Both waves have the same frequency, wavelength, and amplitude but travel in opposite directions. The superposition of these two waves produces a pattern with fixed positions of zero displacement and fixed positions of maximum displacement.

On a stretched string fixed at both ends, a progressive wave travels along the string, reflects at the far end, and the reflected wave travels back. The incident and reflected waves superpose to create a standing wave.

Standing waves can also form in air columns (wind instruments, organ pipes) and with microwaves reflected from a metal plate.

Nodes and Antinodes

Nodes are points on a standing wave where the displacement is always zero. At a node, the two progressive waves always cancel each other out completely (destructive interference at all times).

Antinodes are points where the displacement oscillates between the maximum positive and maximum negative values. At an antinode, the two progressive waves always reinforce each other (constructive interference, alternating between maximum positive and maximum negative displacement).

Key features:

The distance between adjacent nodes is λ/2 (half a wavelength)
The distance between adjacent antinodes is also λ/2
The distance between a node and the nearest antinode is λ/4
All points between two adjacent nodes oscillate in phase with each other
Points on opposite sides of a node oscillate in antiphase (180° phase difference)

Standing Waves vs Progressive Waves

Property	Progressive wave	Standing wave
Energy transfer	Transfers energy in the direction of travel	No net energy transfer
Amplitude	All points have the same amplitude	Amplitude varies from zero (node) to maximum (antinode)
Phase	Phase varies continuously along the wave	All points between adjacent nodes are in phase
Wavelength	Distance between adjacent points in phase	Twice the distance between adjacent nodes
Frequency	All points oscillate at the same frequency	All points oscillate at the same frequency
Formation	Single wave travelling in one direction	Superposition of two waves travelling in opposite directions

Harmonics on a Stretched String

A string fixed at both ends must have nodes at both ends (the fixed points cannot move). The standing wave patterns that satisfy this condition are called harmonics (or modes of vibration).

First harmonic (fundamental, n = 1): The simplest pattern has a node at each end and one antinode in the middle. The string vibrates as a single segment. The length of the string equals half a wavelength:

$L = \frac{\lambda_1}{2} \quad \Rightarrow \quad \lambda_1 = 2L$

The fundamental frequency is:

$f_1 = \frac{v}{\lambda_1} = \frac{v}{2L}$

Second harmonic (n = 2): Two antinodes, three nodes (including the two ends). The string vibrates in two segments.

$L = \lambda_2 \quad \Rightarrow \quad \lambda_2 = L \quad \Rightarrow \quad f_2 = \frac{v}{L} = 2f_1$

Third harmonic (n = 3): Three antinodes, four nodes.

$L = \frac{3\lambda_3}{2} \quad \Rightarrow \quad \lambda_3 = \frac{2L}{3} \quad \Rightarrow \quad f_3 = \frac{3v}{2L} = 3f_1$

General formula:

$f_n = \frac{nv}{2L} = nf_1$

The nth harmonic has n antinodes and (n + 1) nodes (including the end nodes). Every harmonic is an integer multiple of the fundamental frequency.

Harmonic	n	Antinodes	Nodes	Wavelength	Frequency
1st (fundamental)	1	1	2	2L	v/(2L)
2nd	2	2	3	L	v/L
3rd	3	3	4	2L/3	3v/(2L)
4th	4	4	5	L/2	2v/L
nth	n	n	n+1	2L/n	nv/(2L)

Worked example 1: A guitar string is 0.64 m long and the speed of a transverse wave on it is 320 m s⁻¹. Calculate the fundamental frequency and the frequency of the third harmonic.

f₁ = v/(2L) = 320/(2 × 0.64) = 320/1.28 = 250 Hz

f₃ = 3f₁ = 3 × 250 = 750 Hz

Worked example 2: A string vibrates at its second harmonic at 440 Hz (concert A). The string is 0.33 m long. What is the speed of the wave on the string?

For the second harmonic: f₂ = v/L, so v = f₂ × L = 440 × 0.33 = 145.2 m s⁻¹

Standing Waves in Air Columns

Standing waves also form in columns of air, which is the basis of all wind instruments and organ pipes.

Closed Pipe (one end closed, one end open)

At the closed end, air cannot move — this is always a node. At the open end, air can move freely — this is always an antinode.

The simplest pattern (fundamental) has one node and one antinode, with the pipe length equal to one quarter wavelength:

$L = \frac{\lambda_1}{4} \quad \Rightarrow \quad f_1 = \frac{v}{4L}$

A closed pipe produces only odd harmonics: f₁, 3f₁, 5f₁, 7f₁, ...

This gives closed-pipe instruments (like a clarinet) a distinctive, mellow timbre because even harmonics are absent.

Open Pipe (both ends open)

Both ends are antinodes. The simplest pattern has an antinode at each end and a node in the middle, with the pipe length equal to half a wavelength:

$L = \frac{\lambda_1}{2} \quad \Rightarrow \quad f_1 = \frac{v}{2L}$

An open pipe produces all harmonics: f₁, 2f₁, 3f₁, 4f₁, ...

This gives open-pipe instruments (like a flute) a brighter, richer sound.

Pipe type	Boundary conditions	Fundamental λ	Fundamental f	Harmonics present
Closed (one end)	Node + antinode	4L	v/(4L)	Odd only: 1, 3, 5...
Open (both ends)	Antinode + antinode	2L	v/(2L)	All: 1, 2, 3, 4...

Worked example 3: An organ pipe open at both ends is 0.85 m long. The speed of sound is 340 m s⁻¹. Calculate the fundamental frequency and the first three harmonics.

f₁ = v/(2L) = 340/(2 × 0.85) = 340/1.70 = 200 Hz

f₂ = 2f₁ = 400 Hz, f₃ = 3f₁ = 600 Hz

Worked example 4: A closed organ pipe produces a fundamental note of 262 Hz (middle C). Calculate the length of the pipe. (Speed of sound = 340 m s⁻¹)

L = v/(4f₁) = 340/(4 × 262) = 340/1048 = 0.324 m ≈ 32 cm

The next resonant frequency is the 3rd harmonic: 3 × 262 = 786 Hz (no 2nd harmonic in a closed pipe).

Measuring the Speed of Sound Using Standing Waves

A common experiment uses a resonance tube — a tube partially submerged in water, creating a closed pipe of adjustable length. A vibrating tuning fork of known frequency is held above the open end. As the tube is raised out of the water, the air column lengthens. At specific lengths, resonance occurs and a loud sound is heard.

The first resonance occurs when L₁ ≈ λ/4, and the second resonance at L₂ ≈ 3λ/4. The difference gives:

$L_2 - L_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{\lambda}{2}$

So λ = 2(L₂ − L₁), and then v = fλ.

Worked example 5: A tuning fork of frequency 512 Hz is used. The first resonance occurs at a tube length of 16.0 cm and the second at 49.5 cm. Calculate the speed of sound.

λ = 2(L₂ − L₁) = 2(0.495 − 0.160) = 2 × 0.335 = 0.670 m

v = fλ = 512 × 0.670 = 343 m s⁻¹

This is in good agreement with the accepted value of 343 m s⁻¹ at 20 °C.

Real-world application: Standing waves determine the pitch of every stringed and wind instrument. A guitar player pressing a fret changes the effective length of the vibrating string, changing the wavelength and therefore the pitch. Harmonics on a guitar are produced by lightly touching the string at a node position (e.g., the 12th fret for the 2nd harmonic), forcing a specific pattern.

Stationary Waves with Microwaves

A microwave transmitter pointed at a metal reflector creates standing waves. A detector probe moved along the path between the transmitter and reflector detects alternating maxima and minima of intensity. The distance between successive minima (nodes) is λ/2, allowing the microwave wavelength to be measured.

For microwaves of frequency 10 GHz: λ = c/f = 3.0 × 10⁸ / 10 × 10⁹ = 0.030 m = 3.0 cm

The node spacing would be 1.5 cm — easily measurable with a ruler.

Common exam mistake: Students sometimes state that "the amplitude is the same at all points in a standing wave." This is true for progressive waves but false for standing waves. In a standing wave, the amplitude varies continuously from zero at nodes to a maximum at antinodes.

A-Level Deep Dive: Standing Waves

Spec mapping

Edexcel 9PH0 specification Topic 5 — Waves and the Particle Nature of Light, in its sub-strand on stationary (standing) waves, requires students to describe how stationary waves are formed by the superposition of two progressive waves of the same frequency travelling in opposite directions; identify and locate nodes and antinodes; derive and use harmonic frequencies for strings fixed at both ends and for air columns in pipes that are open or closed at one end; and apply $f_n = nv/(2L)$ for strings and open pipes and $f_n = (2n-1)v/(4L)$ for closed pipes (refer to the official Pearson Edexcel specification document for exact wording). The standing-wave material connects backwards to the opening sub-strand on $v = f\lambda$ , and forwards to Topic 6 (further mechanics — SHM and resonance), where every point on a standing wave executes simple harmonic motion at the local amplitude. It is also load-bearing for Core Practical 3 (speed of sound by resonance), which uses the standing-wave node/antinode geometry to extract a wavelength from measured tube lengths. The Edexcel data-and-formulae booklet lists $v = f\lambda$ , but does not list the harmonic series formulae — the relationships $L = n\lambda/2$ (string or open pipe) and $L = (2n-1)\lambda/4$ (closed pipe) must be derived from the boundary conditions on the spot.

Worked example with full mark scheme

Question (8 marks):

A steel string of length $0.65\ \text{m}$ is fixed at both ends and plucked. The fundamental frequency is measured as $256\ \text{Hz}$ .

(a) Calculate the wavelength of the fundamental standing wave on the string. (2)

(b) Calculate the speed of transverse waves on the string. (2)

(c) State the frequencies of the second and third harmonics, and sketch the displacement pattern (in words) of the third harmonic, naming the number of nodes and antinodes (excluding the fixed ends as nodes). (4)

Solution with mark scheme:

(a) Step 1 — apply the boundary condition for a string fixed at both ends.

Both ends are nodes, so the lowest mode has a single antinode in the middle and $L = \lambda/2$ .

$\lambda = 2L = 2 \times 0.65 = 1.30\ \text{m}$

M1 — recognising $L = \lambda/2$ for the fundamental of a doubly fixed string. Common slip: writing $L = \lambda$ , halving the wavelength.

A1 — $\lambda = 1.30\ \text{m}$ .

(b) Step 1 — apply the wave equation.

$v = f\lambda = 256 \times 1.30 = 333\ \text{m s}^{-1}$

M1 — substitution into $v = f\lambda$ .

A1 — $v = 333\ \text{m s}^{-1}$ (or $330$ to 3 sf). A unit error (" $333\ \text{m}$ ") loses the A1.

Allowed wavelengths $\lambda_n = 2L/n$ , so $f_n = nf_1$ . Hence $f_2 = 512\ \text{Hz}$ and $f_3 = 768\ \text{Hz}$ .

M1 — recognising the integer-multiple harmonic series.

A1 — both values correct.

Step 2 — describe the third-harmonic mode.

$L = 3\lambda/2$ , so the string is divided into three antinodal loops: two interior nodes (plus two fixed-end nodes) and three antinodes.

M1 — identifying three loops.

A1 — explicit node and antinode counts.

Total: 8 marks (M4 A4).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A glass tube of length $0.34\ \text{m}$ , closed at the bottom and open at the top, has a small loudspeaker held above the open end. The loudspeaker frequency is increased slowly from $200\ \text{Hz}$ . The first loud resonance is heard at $250\ \text{Hz}$ and the second loud resonance at $750\ \text{Hz}$ .

(a) Calculate the speed of sound in the air column from the first-resonance data. (2)

(b) Show that the second resonance corresponds to the third harmonic of the closed pipe, and use it to obtain a second value for the speed of sound. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1) — applying $L = \lambda/4$ at the first resonance, giving $\lambda_1 = 4L = 1.36\ \text{m}$ .
A1 (AO1) — $v = f_1\lambda_1 = 250 \times 1.36 = 340\ \text{m s}^{-1}$ .

(b)

M1 (AO2) — recognising that the closed-pipe series is $f_n = (2n-1)f_1$ , with $n = 2$ giving $f = 3f_1 = 750\ \text{Hz}$ , matching the observed second resonance, and hence $\lambda = 4L/3 = 0.453\ \text{m}$ .
A1 (AO1) — $v = 750 \times 0.453 = 340\ \text{m s}^{-1}$ (consistent with (a)).

(c)

M1 (AO3) — naming an end-correction effect: the antinode at the open end sits slightly above the rim, so the effective acoustic length exceeds the geometric length, producing a small systematic underestimate of $v$ when using $L$ as if it were exact.
A1 (AO3) — concluding that the higher-harmonic value is generally more reliable because the fractional contribution of the end correction to the acoustic length decreases for shorter wavelengths.

Total: 6 marks split AO1 = 3, AO2 = 1, AO3 = 2. This is typical of Edexcel resonance questions: the calculation is procedural, but the AO3 marks come from explaining systematic error in the geometric-vs-acoustic length distinction.

Synoptic links

Connects to:

Topic 5 — superposition and interference: standing waves are the special case of two-source superposition in which a forward-travelling and a reflected wave of identical frequency and amplitude overlap. Every node is a permanent destructive-interference point, every antinode a permanent constructive one.
Topic 6 — SHM and resonance: every point between nodes executes SHM at the standing-wave frequency, with local amplitude $A(x) = 2A_0\sin(2\pi x/\lambda)$ . The string resonates at $f_n$ for the same reason a mass-on-spring resonates: the driving frequency matches a normal mode.
Music acoustics: the harmonic series $f_n = nf_1$ for a string determines musical timbre. Plucking at the midpoint suppresses even harmonics (the midpoint is a node for $f_2, f_4, \ldots$ ); plucking near the bridge excites a richer mix — the basis of "warm" vs "bright" tone.