Stationary Waves — OCR A Level Physics Revision
Module 4 · Waves

Stationary Waves

Specification: OCR A H556  |  Section: 4.4.4  |  Teaching time: ~3 hours

By the end of this topic you should be able to…
Part 1

Formation of Stationary Waves

A stationary wave (or standing wave) is formed when two progressive waves of the same frequency and amplitude travelling in opposite directions superpose (meet and combine).

How Stationary Waves Form

The most common way to form a stationary wave is by reflection. When a wave meets a fixed boundary, it reflects back along the same path. The incident wave and reflected wave then superpose.

For a stationary wave to form:

  • Both waves must have the same frequency (and therefore same wavelength)
  • Both waves must have the same amplitude
  • The waves must travel in opposite directions
  • They must be coherent (constant phase relationship)
💡 Key Idea

In a stationary wave, energy is stored rather than transferred. The wave appears to "stand still" — the positions of maximum and minimum displacement don't move along the medium. This is fundamentally different from progressive waves, which transfer energy.

Superposition and the Resultant Wave

At any point along the wave, the resultant displacement is the sum of the displacements of the two progressive waves. This is the principle of superposition.

Part 2

Nodes and Antinodes

The most distinctive feature of stationary waves is the pattern of nodes and antinodes.

Nodes

A node is a point along the stationary wave where the displacement is always zero. At a node:

  • The two progressive waves are always in antiphase (180° apart)
  • They cancel by destructive interference
  • The amplitude of oscillation is zero
  • Particles at nodes don't move at all

Antinodes

An antinode is a point where the displacement varies by the maximum amount. At an antinode:

  • The two progressive waves are always in phase
  • They add by constructive interference
  • The amplitude of oscillation is maximum (2A, where A is the amplitude of each progressive wave)
  • Particles at antinodes oscillate with the largest amplitude
Node-Antinode Separation Distance between adjacent nodes = λ/2
Distance between adjacent antinodes = λ/2
Distance between a node and adjacent antinode = λ/4
⚠️ Common Misconception

Nodes are NOT points where "nothing happens" — they're points of zero displacement. The medium on either side of a node is oscillating, but the node itself remains stationary. Also, all points between two nodes oscillate in phase with each other, but in antiphase with points in the next "loop".

Phase Relationships in Stationary Waves

  • All points between two adjacent nodes oscillate in phase
  • Points in adjacent loops (separated by a node) are in antiphase (180° out of phase)
  • This is very different from progressive waves, where phase varies continuously with position
Interactive

Stationary Wave Simulator

Stationary Wave Formation
1st
Harmonic: 1st (fundamental)
Wavelength: 2L
Frequency: f₀
Nodes: 2

Adjust the harmonic slider to see different modes of vibration. The blue and green waves are the two progressive waves travelling in opposite directions; the yellow wave is their superposition (the stationary wave). Red dots mark nodes, blue dots mark antinodes.

Part 3

Stationary vs Progressive Waves

Understanding the differences between stationary and progressive waves is essential for exam questions.

Property Progressive Wave Stationary Wave
Energy transfer Energy is transferred in the direction of travel No energy transfer (energy is stored)
Amplitude Same for all particles (assuming no damping) Varies with position: zero at nodes, maximum at antinodes
Wavelength Distance between two consecutive points in phase Twice the distance between adjacent nodes (λ = 2 × node spacing)
Phase Changes continuously with position All points in one "loop" are in phase; adjacent loops in antiphase
Frequency All particles oscillate at same frequency All particles (except nodes) oscillate at same frequency
Wave pattern Wave shape moves through medium Wave pattern appears stationary; only amplitude oscillates
Worked Example
A stationary wave on a string has a distance of 0.75 m between the first and third nodes. Calculate the wavelength of the wave.
Distance between adjacent nodes = λ/2
From 1st to 3rd node, there are 2 node separations
Therefore: 2 × (λ/2) = 0.75 m ✓
λ = 0.75 m ✓
Check your understanding

Knowledge Check

1
State what is meant by a node in a stationary wave.
1 mark
  • A point where the displacement is always zero / amplitude is zero ✓
  • Or: A point where the two waves interfere destructively ✓
2
A stationary wave has a wavelength of 1.2 m. Calculate the distance between adjacent nodes.
1 mark
  • Distance between nodes = λ/2 = 1.2/2 = 0.60 m ✓
3
Explain why stationary waves do not transfer energy.
2 marks
  • The two progressive waves travel in opposite directions ✓
  • They carry equal amounts of energy in opposite directions, so net energy transfer is zero ✓
4
Describe how the phase varies along a stationary wave compared to a progressive wave.
2 marks
  • Progressive: phase changes continuously with position ✓
  • Stationary: all points between adjacent nodes are in phase; adjacent loops are in antiphase ✓
Part 4

Harmonics on Stretched Strings

When a string is fixed at both ends and made to vibrate, it can only vibrate at certain frequencies called harmonics. These are determined by the string's length, tension, and mass per unit length.

Boundary Conditions for Strings

At each fixed end, there must be a node (the string cannot move). This constrains which wavelengths are possible.

The Harmonics

The fundamental mode (1st harmonic) is the lowest frequency at which the string can vibrate. Higher harmonics are integer multiples of this frequency.

1st Harmonic (Fundamental)
λ = 2L   f = f₀
2nd Harmonic
λ = L   f = 2f₀
3rd Harmonic
λ = 2L/3   f = 3f₀
General Harmonic for Stretched String λ_n = 2L/n    f_n = n × f₀ = n/(2L) × √(T/μ)

Where:

  • L = length of string
  • n = harmonic number (1, 2, 3, ...)
  • T = tension in string
  • μ = mass per unit length
Worked Example
A guitar string of length 0.65 m vibrates at its fundamental frequency of 220 Hz. Calculate:
(a) The wavelength of the wave on the string
(b) The speed of waves on the string
(c) The frequency of the 3rd harmonic
For fundamental: λ = 2L = 2 × 0.65 = 1.30 m ✓
v = fλ = 220 × 1.30 = 286 m s⁻¹ ✓
3rd harmonic: f₃ = 3 × 220 = 660 Hz ✓
Part 5

Stationary Waves in Air Columns

Stationary waves can also form in air columns inside pipes. The behaviour depends on whether the pipe is open or closed at each end.

Boundary Conditions for Air Columns

  • Open end: Must be an antinode (air can oscillate freely, pressure is atmospheric)
  • Closed end: Must be a node (air cannot move, pressure varies maximum)

Pipe Open at Both Ends

Both ends are antinodes. The harmonics are the same as for a string:

Open Pipe Harmonics λ_n = 2L/n    f_n = n × v/(2L)    (n = 1, 2, 3, ...)

All harmonics are present (n = 1, 2, 3, 4, ...).

Pipe Closed at One End

One node (closed end) and one antinode (open end). Only odd harmonics are present:

Closed Pipe Harmonics λ_n = 4L/n    f_n = n × v/(4L)    (n = 1, 3, 5, ... odd only)
📋 Key Exam Point

A closed pipe has only odd harmonics (1st, 3rd, 5th, ...). The fundamental wavelength is λ = 4L (four times the pipe length!), compared to λ = 2L for an open pipe. This is because the shortest standing wave has a node at one end and antinode at the other — that's a quarter wavelength!

Worked Example
A pipe closed at one end has length 0.34 m. The speed of sound in air is 340 m s⁻¹.
(a) Calculate the fundamental frequency.
(b) What is the frequency of the next harmonic that can be formed?
(c) Compare this to an open pipe of the same length.
Fundamental (closed): f₁ = v/(4L) = 340/(4 × 0.34) = 250 Hz ✓
Next harmonic is 3rd (odd only): f₃ = 3 × 250 = 750 Hz ✓
Open pipe fundamental: f₁ = v/(2L) = 340/(2 × 0.34) = 500 Hz (higher than closed pipe) ✓
Part 6

Measuring the Speed of Sound

The resonance tube experiment is a required practical technique for determining the speed of sound in air.

Method

  1. Place a tuning fork of known frequency f above a tube partially submerged in water
  2. Raise or lower the tube to change the effective air column length
  3. Find the position where the sound is loudest (resonance) — this corresponds to a stationary wave
  4. Measure the length L of the air column at resonance

Theory

For a tube closed at one end, the first resonance occurs when:

First Resonance Position L + e = λ/4

Where e is an "end correction" (typically 0.6 × tube radius) accounting for the fact that the antinode is slightly outside the tube.

The second resonance occurs at:

Second Resonance Position L₂ + e = 3λ/4

Eliminating End Correction

By measuring both resonance positions and subtracting:

Wavelength from Two Resonances L₂ - L₁ = 3λ/4 - λ/4 = λ/2  ∴ λ = 2(L₂ - L₁)

Then: v = fλ

💡 Practical Tip

Always measure multiple resonance positions and plot a graph of resonance position against (2n-1) for the nth resonance. The gradient gives λ/4, eliminating systematic errors from end correction.

Exam Practice

Exam-Style Questions

1
(a) Explain how a stationary wave is formed on a stretched string fixed at both ends. [3 marks]

(b) A string of length 0.60 m is fixed at both ends. When it vibrates in its second harmonic mode, the frequency is 440 Hz. Calculate the speed of waves on the string. [3 marks]

(c) Determine the fundamental frequency of this string. [1 mark]
7 marks

(a) [3]:

  • Wave travels along string and reflects at fixed end ✓
  • Incident and reflected waves superpose ✓
  • For certain frequencies, nodes form at fixed ends with antinodes between ✓

(b) [3]:

  • 2nd harmonic: λ = L = 0.60 m ✓
  • v = fλ = 440 × 0.60 ✓
  • v = 264 m s⁻¹ ✓

(c) [1]:

  • f₁ = f₂/2 = 440/2 = 220 Hz ✓
2
A pipe of length L is closed at one end and open at the other. The speed of sound in air is 330 m s⁻¹.

(a) Draw the stationary wave pattern for the fundamental mode of vibration, labelling the node and antinode. [2 marks]

(b) Show that the fundamental frequency is f₁ = 330/(4L). [2 marks]

(c) The pipe has length 0.825 m. Calculate the fundamental frequency and the frequency of the next harmonic present. [2 marks]

(d) Explain why the 2nd harmonic (n = 2) cannot exist in this pipe. [2 marks]
8 marks

(a) [2]:

  • Node at closed end, antinode at open end ✓
  • Correct shape (quarter wavelength) ✓

(b) [2]:

  • Fundamental: L = λ/4, so λ = 4L ✓
  • f = v/λ = v/(4L) = 330/(4L) ✓

(c) [2]:

  • f₁ = 330/(4 × 0.825) = 100 Hz ✓
  • Next harmonic is 3rd: f₃ = 3 × 100 = 300 Hz ✓

(d) [2]:

  • 2nd harmonic would require both ends to be same type (both nodes or both antinodes) ✓
  • Closed pipe must have node at one end, antinode at other — only odd harmonics satisfy this ✓
3
A student uses a resonance tube to determine the speed of sound. A tuning fork of frequency 512 Hz is sounded above the tube.

(a) The first resonance is heard when the air column length is 16.0 cm. A second resonance is heard when the length is 49.2 cm. Calculate the wavelength of the sound. [2 marks]

(b) Calculate the speed of sound. [1 mark]

(c) Explain the advantage of using two resonance positions rather than one. [2 marks]
5 marks

(a) [2]:

  • L₂ - L₁ = λ/2 ✓
  • λ = 2 × (49.2 - 16.0) × 10⁻² = 2 × 0.332 = 0.664 m ✓

(b) [1]:

  • v = fλ = 512 × 0.664 = 340 m s⁻¹ ✓

(c) [2]:

  • Using two positions eliminates the end correction (e cancels out) ✓
  • Results in more accurate determination of wavelength/speed of sound ✓
4
(a) State two differences between stationary waves and progressive waves. [2 marks]

(b) A stationary wave is formed on a string. The distance between the first node and the fifth node is 1.60 m. Calculate the wavelength of the wave. [2 marks]

(c) All points between two adjacent nodes oscillate in phase. Explain what this means. [2 marks]
6 marks

(a) [2]: Any two:

  • Progressive transfers energy; stationary stores energy ✓
  • Progressive has constant amplitude; stationary amplitude varies with position ✓
  • Progressive phase changes continuously; stationary phase is constant between nodes ✓

(b) [2]:

  • From 1st to 5th node = 4 node separations = 4 × λ/2 = 2λ ✓
  • 2λ = 1.60 m, so λ = 0.80 m ✓

(c) [2]:

  • All particles between nodes reach maximum displacement at the same time ✓
  • They all pass through equilibrium at the same time (zero phase difference) ✓
5
A flute can be modelled as a pipe open at both ends. A particular flute has an effective length of 0.67 m.

(a) Calculate the fundamental frequency of this flute. Speed of sound = 330 m s⁻¹. [2 marks]

(b) The flautist can produce notes at higher frequencies by overblowing. State the frequencies of the next two harmonics that can be produced. [2 marks]

(c) A clarinet is closed at one end. Compare the harmonics available on a clarinet to those on a flute of the same length. [3 marks]
7 marks

(a) [2]:

  • Open pipe fundamental: λ = 2L = 1.34 m ✓
  • f₁ = v/λ = 330/1.34 = 246 Hz (or using f = v/2L = 330/(2×0.67)) ✓

(b) [2]:

  • 2nd harmonic: f₂ = 2 × 246 = 492 Hz ✓
  • 3rd harmonic: f₃ = 3 × 246 = 738 Hz ✓

(c) [3]:

  • Clarinet (closed pipe) has only odd harmonics; flute (open pipe) has all harmonics ✓
  • Clarinet fundamental: f₁ = v/4L = 123 Hz (half the flute's fundamental) ✓
  • Clarinet overtones: 123 Hz, 369 Hz, 615 Hz... (odd multiples only) vs flute: 246 Hz, 492 Hz, 738 Hz... ✓

Topic Summary

Formation

Two progressive waves of same frequency and amplitude, travelling in opposite directions, superpose to form a stationary wave.

Nodes & Antinodes

Nodes: zero amplitude (destructive interference). Antinodes: maximum amplitude (constructive interference). Separation = λ/2.

Strings

Fixed ends are nodes. λ_n = 2L/n. All harmonics present. f_n = n × f₀.

Air Columns

Open end = antinode. Closed end = node. Open pipe: all harmonics. Closed pipe: odd harmonics only.

Equations to Know

Node spacing = λ/2
String: λ_n = 2L/n
Open pipe: λ_n = 2L/n
Closed pipe: λ_n = 4L/n (n odd)
v = fλ
f_n = n × f₀