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Standing Waves in Air Columns (Wavelength of Sound)

Key idea: Identify displacement vs pressure nodes/antinodes in air columns and determine the wavelength of sound using standing waves (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Explain standing-wave formation, nodes, antinodes and energy transfer.
  • Apply boundary conditions to standing waves on stretched strings.
  • Analyse displacement and pressure patterns in resonant air columns and determine sound wavelength.

1. Definitions (Must Know)

  • Standing wave: a wave pattern produced by superposition of two waves of the same frequency travelling in opposite directions.
  • Displacement node: a point where particle displacement is always zero.
  • Displacement antinode: a point where particle displacement amplitude is maximum.
  • Pressure node: a point where pressure variation is zero.
  • Pressure antinode: a point where pressure variation amplitude is maximum.
  • Air column: a column of air in a tube, which can support standing sound waves.

2. Key Ideas (What Earns Marks)

  • Sound in air is longitudinal:
    • particle displacement and pressure variations are along the tube.
  • End conditions (important):
    • Open end:
      • displacement antinode
      • pressure node
    • Closed end:
      • displacement node
      • pressure antinode
  • Allowed wavelengths (tube length L):
    • Open–open tube (both ends open):
      • L = nλ/2 (n = 1,2,3,…)
    • Closed–open tube (one end closed):
      • L = ((2n-1)λ)/4 (n = 1,2,3,…)
  • Finding wavelength from resonance lengths:
    • successive resonances differ by:
      • Δ L = λ/2

3. Detailed Explanations

A. Why open and closed ends differ

  • At a closed end, air cannot move, so displacement must be zero → displacement node.
  • At an open end, air can move freely, so displacement is maximum → displacement antinode.

Pressure variation is “opposite”:

  • where displacement is a node, pressure is an antinode
  • where displacement is an antinode, pressure is a node

B. Open–open tube

Both ends are displacement antinodes, so a whole number of half-wavelengths fits: L = nλ/2

Fundamental (n = 1): L = λ/2.

C. Closed–open tube

Closed end is a displacement node, open end is a displacement antinode, so an odd number of quarter-wavelengths fits: L = ((2n-1)λ)/4

Fundamental (n = 1): L = λ/4.

Only odd harmonics occur for the closed–open tube.

Resonance lengths for air columns (example)

Resonant lengths for open–open and closed–open tubes for the same wavelength.

Scroll across the graph to read all labels.

Resonant lengths for open–open and closed–open tubes for the same wavelength.Resonant lengths for open–open and closed–open tubes for the same wavelength.
Open–open tubes have all harmonics; closed–open tubes have only odd harmonics. In both cases, successive resonance lengths differ by Δ L = λ/2.
Open full-size graph
View figure data
Values for Resonance lengths for air columns (example)
Mode number, n (unitless)Open–open: L = nλ/2 (λ = 0.80 m)Closed–open: L = (2n−1)λ/4 (λ = 0.80 m)
10.40.2
20.80.6
31.21
41.61.4

D. Measuring wavelength using resonance

In a resonance tube experiment, you adjust the air-column length until the sound is loudest (resonance).

If you find two successive resonance lengths L₁ and L₂ for the same frequency: λ = 2(L₂-L₁)

Then wave speed: v = fλ

4. Common Mistakes

  • Saying “node at open end” for displacement (it is an antinode).
  • Mixing up displacement and pressure nodes/antinodes.
  • Using L = λ/2 for a closed–open tube fundamental (it should be L = λ/4).
  • Using v = fλ with inconsistent units (Hz with cm).

5. Exam Tips

  • If a question mentions “loudest sound” in a tube: it is a standing-wave resonance condition.
  • State end conditions explicitly before writing the wavelength relation.
  • When using two resonance lengths, quote Δ L = λ/2 first (easy marks).

6. Worked Examples

Modelled example 1

Wavelength from two resonance lengths

Core

Problem

A closed-end resonance tube driven at 512 Hz has successive resonances at 0.16 m and 0.49 m. Find wavelength and sound speed.
Study the worked solution

  1. Use resonance spacing

    Method

    The length difference is half a wavelength.

    Reason

    Successive modes add one half-wavelength while retaining the same end conditions.

    Working

    L₂-L₁ = λ/2

  2. Find wavelength

    Method

    λ = 0.66 m.

    Reason

    Double the measured difference.

    Working

    λ = 2(0.49-0.16) = 0.66 m

  3. Find speed

    Method

    v = 3.38 × 10² m s⁻¹.

    Reason

    Use v = fλ with the tuning-fork frequency.

    Working

    v = (512)(0.66) = 3.38 × 10² m s⁻¹

Guided practice 2

Fundamental length for a closed–open tube

About 3 min

Problem

Sound of wavelength 0.80 m forms the fundamental in a tube closed at one end. Find the air-column length.

Try this before viewing the solution

Unit: m

Hints

Hint 1: apply unlike end conditions
The mode runs from a displacement node at the closed end to an antinode at the open end.
View solution step by step

  1. Identify the geometry

    Method

    The tube contains one quarter-wavelength.

    Reason

    A node-to-nearest-antinode separation is λ/4.

    Working

    L = λ/4

  2. Calculate

    Method

    L = 0.20 m.

    Reason

    Divide the wavelength by four.

    Working

    L = 0.80/4 = 0.20 m

Common misconception 3

Identifying nodes/antinodes

Find and correct the mistake

Learner claim

For a closed–open tube, a learner labels the open end as both a displacement node and a pressure node. Diagnose the labels at both ends.

Try this before viewing the solution

View solution step by step

  1. Label the open end

    Method

    The open end is a displacement antinode and pressure node.

    Reason

    Air moves freely while pressure stays close to atmospheric.

    Working

    open: displacement A; pressure N

  2. Label the closed end

    Method

    The closed end is a displacement node and pressure antinode.

    Reason

    Air cannot move through the wall while pressure variation is largest.

    Working

    closed: displacement N; pressure A

  3. State the pairing

    Method

    Displacement nodes correspond to pressure antinodes, and conversely.

    Reason

    Particle motion and pressure variation are spatially offset.

    Working

    displacement N ↔ pressure A

Examiner practice 4

Open–open tube fundamental

3 marks

Examination question

An open–open tube of length 0.85 m resonates at its fundamental. Find the sound wavelength. [3 marks]

Try this before viewing the solution

Unit: m

View solution step by step

  1. State the ends

    1 mark

    Method

    Both open ends are displacement antinodes.

    Reason

    The fundamental spans adjacent antinodes with one node between.

    Working

    open–open: A–N–A

  2. Relate length and wavelength

    1 mark

    Method

    L = λ/2.

    Reason

    Adjacent antinodes are separated by half a wavelength.

    Working

    λ = 2L

  3. Calculate

    1 mark

    Method

    λ = 1.70 m.

    Reason

    Double the tube length.

    Working

    λ = 2(0.85) = 1.70 m

Challenge 5

Finding frequency from resonance spacing

Minimal support

Independent transfer

Successive resonance lengths differ by 0.18 m. Find the wavelength and then the frequency when sound speed is 340 m s⁻¹.

Try this before viewing the solution

Hints

Hint 1: convert spacing before wave speed
Successive resonance spacing is λ/2, not λ.
View solution step by step

  1. Recover wavelength

    Method

    λ = 0.36 m.

    Reason

    Successive resonant lengths differ by half a wavelength.

    Working

    λ = 2(0.18) = 0.36 m

  2. Use the wave relation

    Method

    f = 9.44 × 10² Hz.

    Reason

    Rearrange v = fλ after obtaining the full wavelength.

    Working

    f = 340/0.36 = 9.44 × 10² Hz

7. Mind Stretchers

Mind stretcher 1: Which harmonics exist?Extension

Explain why a closed–open tube has only odd harmonics.

Show Answer

The boundary conditions are different at the two ends (node at closed end, antinode at open end). This requires the length to contain an odd number of quarter-wavelengths, L = (2n-1)λ/4, which corresponds to odd harmonics only.

Mind stretcher 2: Why is pressure a node at an open end?Extension

Explain (qualitatively) why the open end of a tube is a pressure node for a standing sound wave.

Show Answer

At an open end, the air is in contact with the atmosphere, so the pressure must stay close to atmospheric pressure.

That means the pressure variation is minimal there (approximately zero), so it is a pressure node.

Mind stretcher 3: Simulation Bridge: Standing Wave ExplorerExtension

Concept Explorer: Standing Wave Explorer

Toggle string and air-column boundary conditions, change harmonic mode, and test wavelength-frequency relations.

BetaA LevelWavesBest for: A Level waves and superposition
  • Boundary Conditions
  • Harmonics
  • Resonance Spacing
  • f–λ–v Links

Open the full interactive simulation on its own page

Use the standalone simulation page for the live controls, SVG scene, run modes, and scoring flow.

Open Interactive Simulation

The lesson stays lightweight and links out to the dedicated simulation page.

Investigate air columns in the Standing Wave Explorer.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027