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Waves · Subtopic 2 of 6

Stationary Waves

When two identical progressive waves travel in opposite directions, they superpose to form a stationary wave. Nodes (zero amplitude) and antinodes (maximum amplitude) stay in fixed places, λ2\tfrac\lambda2 apart. No energy is carried along. On a string fixed at both ends the fundamental has λ=2L\lambda=2L; in a pipe closed at one end, λ=4L\lambda=4L.

Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH12AQA 7408 Paper 1≈ 40 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Explain how a stationary wave forms by superposition.
  • Identify nodes and antinodes and use node spacing =λ2=\tfrac\lambda2.
  • Find the harmonics of strings and air columns.
  • Measure the wavelength of sound or microwaves using stationary waves.
Pause and think

A guitar string is fixed at both ends. When plucked, which wavelengths can it vibrate with, and why not any wavelength?

Section 2 of 14

Core idea

Start hereA guitar string

Pluck a guitar string and waves race to each end, reflect, and overlap with themselves. Only certain wavelengths fit exactly, producing a pattern that seems to stand still. Those wavelengths decide the notes the string can play.

The big idea

Two equal waves in opposite directions → stationary wave: nodes and antinodes; adjacent nodes λ2\frac\lambda2 apart; no net energy transfer. Strings: L=nλ2L=\frac{n\lambda}2. Closed pipe: odd quarter-wavelengths.

1. Formation

A stationary wave on a string: the shape oscillates, nodes stay still
2468−2−1120position along stringdisplacementnodeantinode
Between two nodes every point moves in phase; on opposite sides of a node they are in antiphase.
  • Between adjacent nodes, all points oscillate in phase.
  • Amplitude varies with position.

2. Strings

Allowed frequencies on a string
  1. L=nλ2L=n\frac\lambda2
    Nodes at both fixed ends: a whole number of half-wavelengths.
  2. fn=vλn=nv2Lf_n=\frac v{\lambda_n}=\frac{nv}{2L}
    v=fλv=f\lambda.
Worked exampleA 0.80 m string

Wave speed 120 m s−1^{-1}.

Goal
Fundamental and harmonics.
Strategy
n=1n=1, then multiples.
  1. λ=2L=1.6 m\lambda=2L=1.6\text{ m}
    Fundamental.
  2. f1=1201.6=75 Hzf_1=\frac{120}{1.6}=75\text{ Hz}
    f=v/λf=v/\lambda.
  3. 150, 225,… Hz150,\ 225,\ldots\text{ Hz}
    All multiples.
Answer7575 Hz, 150150 Hz, 225225 Hz, …
✓ Checknv2L\frac{nv}{2L} ✓.

3. Air columns

Worked exampleResonance tube, 500 Hz fork

Resonances at 0.16 m and 0.50 m.

Goal
Speed of sound.
Strategy
Successive resonances differ by λ2\frac\lambda2.
  1. 0.50−0.16=0.34=λ20.50-0.16=0.34=\frac\lambda2
    Subtract: end correction cancels.
  2. λ=0.68 m\lambda=0.68\text{ m}
    Double.
  3. v=500×0.68=340 m s−1v=500\times0.68=340\text{ m s}^{-1}
    fλf\lambda.
Answer340340 m s−1^{-1}.
✓ CheckFirst length 0.16≈λ4=0.170.16\approx\frac\lambda4=0.17: end correction 0.010.01 m ✓.
Common misconceptionNode spacing Adjacent nodes are λ2\frac\lambda2 apart, not λ\lambda.
Pause and think

Closed pipe: which harmonics?

Summary

  • Nodes λ/2\lambda/2 apart.
  • Strings L=nλ/2L=n\lambda/2; closed pipes odd quarters.
Key words
Node
Zero amplitude.
Antinode
Maximum amplitude.
Third harmonic on a string fixed at both ends
0.20.40.60.811.2−1−0.50.510position along string (m)displacementone extremehalf a period laternodenodenodenodeantinode
Nodes are half a wavelength apart. Here L=32λL=\tfrac32\lambda, so λ=0.80\lambda=0.80 m. Points between neighbouring nodes move in phase.
HistoryChladni's singing plates. In 17871787 Ernst Chladni sprinkled sand on metal plates and bowed their edges. The sand collected along the nodal lines, forming beautiful patterns. Napoleon was so impressed he paid for Chladni's book to be translated into French.
Fun factThe Tacoma Narrows Bridge. In 19401940 wind set up a twisting stationary-wave pattern in the Tacoma Narrows Bridge in the USA. A line down the middle of the deck stayed still (a node) while the edges heaved up and down by metres, until the bridge tore itself apart.
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