Superposition and standing-wave experiments

Key idea: H2 Physics lessons on progressive-wave models, standing waves, interference, diffraction and resolution.

  • GCE A-Level H2 Physics 2027

Learn the idea

Big question: How can two travelling waves produce fixed nodes?

A standing wave forms when coherent waves of the same frequency and similar amplitude travel in opposite directions. Nodes remain at zero amplitude; antinodes have maximum amplitude, and neighbouring nodes are λ/2 apart. Investigate this with a stretched string, air column or microwave setup by finding resonant patterns and measuring several node spacings.

Build the pattern from two travelling waves

When equal waves of the same frequency and amplitude travel in opposite directions, their displacements add. Fixed nodes always have zero resultant displacement; antinodes midway between them have maximum amplitude. There is no net energy transfer along an ideal standing wave.

Adjacent nodes are λ/2 apart and a node-to-neighbouring-antinode distance is λ/4. Particles within one loop oscillate in phase; particles in neighbouring loops are in antiphase. The pattern is not a single frozen sinusoid—it changes shape through each cycle.

Check your understanding: Two adjacent nodes are 0.18 m apart. What is the wavelength?

λ = 2(0.18) = 0.36 m.

Apply the correct boundary condition

A stretched string fixed at an end has a displacement node there. An air column closed at one end has an air-displacement node but a pressure antinode; an open end has an air-displacement antinode and approximately a pressure node.

Standing waves can be demonstrated on a driven string, with microwaves reflected from a metal plate, or with resonance in an air column. The measurable quantity differs: string displacement, detector signal for microwave field intensity, or loudness/pressure response in air.

Check your understanding: What air-displacement condition occurs at the closed end of a tube?

A displacement node; the corresponding pressure variation is an antinode.

Standing-wave nodes, antinodes and air-column boundariesTwo opposite string profiles share fixed nodes and opposite antinode displacements. Beside them, open and closed pipe ends are labelled with displacement and pressure boundary conditions.Stretched string: two instants half a cycle apartnodenodenodenodenodeλ/2antinodeantinodeAir column endsopen enddisplacement antinode · pressure nodeclosed end: displacement node · pressure antinode
Scroll diagram horizontally to read all labels.
Neighbouring nodes are λ/2 apart and a node is λ/4 from its nearest antinode. In an air column, an open end is a displacement antinode and pressure node; a closed end is the reverse.

Key ideas to keep

  • A standing wave does not carry net energy along the pattern.
  • All points between adjacent nodes oscillate in phase; neighbouring loops are in antiphase.
  • One node-to-antinode spacing is λ/4.

Worked example

Find sound speed from successive resonances

Question: A tuning fork produces successive resonances in a tube closed at one end at lengths 0.165 m and 0.505 m. Its frequency is 500 Hz. Find wavelength and wave speed.

  1. Step 1: Use the separation, not either absolute length

    Why: The same end correction affects both resonance lengths and cancels in their difference.

    Working: ΔL = 0.505 − 0.165 = 0.340 m.

  2. Step 2: Relate successive modes

    Why: Successive closed-tube resonances differ by half a wavelength.

    Working: λ/2 = 0.340 m, so λ = 0.680 m.

  3. Step 3: Find speed

    Why: Frequency is fixed by the source.

    Working: v = fλ = 500(0.680) = 340 m s⁻¹.

Answer: The wavelength is 0.680 m and sound speed is 340 m s⁻¹.

Check: Using a difference removes the unknown end correction, which is why successive resonances are preferable.

Question

A pipe closed at one end has length 0.850 m and sound speed 340 m s⁻¹. Find its fundamental and next allowed frequency, identify end conditions, and state how resonance can measure wavelength.

Check the worked solution

Closed end: displacement node/pressure antinode; open end: displacement antinode/pressure node. For the fundamental L = λ/4, so λ = 3.40 m and f = 100 Hz. Only odd harmonics occur, so the next is 300 Hz. Successive resonant lengths differ by λ/2, allowing λ and then v = fλ to be found.

Practise with support

Try this

Successive resonant lengths of an air column differ by 0.170 m at 1000 Hz. Find sound wavelength and speed, and state the pressure/displacement conditions at an open end.

Hint: Successive resonances are separated by λ/2.

Check your answer

λ/2 = 0.170 m, so λ = 0.340 m and v = fλ = 340 m s⁻¹. An open end is approximately a pressure node and displacement antinode.

Practise independently

Your turn

Compare how a microwave probe, a vibrating string and a resonance tube reveal standing waves, including what is measured and the relevant node or antinode.

Check your answer

A microwave probe moved through the field records alternating intensity minima and maxima. A driven stretched string shows stationary displacement nodes and antinodes at resonant frequencies. A resonance tube shows sound maxima at allowed air-column lengths; the closed end is a displacement node/pressure antinode and the open end the reverse. Adjacent node or resonance spacing determines λ/2.

Common mistakes

Common mistake

An open end is both a displacement node and a pressure node.

What is wrong with this reasoning?

Show better thinking

An open end is approximately a displacement antinode and pressure node. A closed end is a displacement node and pressure antinode.

Common mistake

Superposition means adding wave intensities point by point.

What is wrong with this reasoning?

Show better thinking

Add signed instantaneous displacements first. Intensity depends on the resulting amplitude.

Exam guidance

On a standing-wave sketch, mark nodes and antinodes and state the boundary conditions.

Exam-style practice [8 marks]

Describe an experiment using a vibrating string to demonstrate standing waves and determine wave speed. Explain the pattern, measurements and one way to improve reliability.

Plan before you answer

  • Name how counter-propagating waves arise.
  • Measure node spacing and frequency.
  • Explain repeated measurements and fixed conditions.
Mark your answer and compare the model

Marking points

Tick each point only if your answer states it clearly.

Model answer

Drive a taut string sinusoidally and arrange reflection at its far end. At resonant frequencies, incident and reflected waves of equal frequency travel oppositely and superpose to form fixed nodes and antinodes. Measure the distance across several node intervals, divide to obtain mean node spacing and use λ = 2d. With the generator frequency f, calculate v = fλ. Measuring many intervals and repeating at fixed tension reduces percentage uncertainty.

Check what stayed with you

Recall question 1

How far apart are adjacent nodes?

Check the answer

λ/2.

Recall question 2

What is the phase relation of neighbouring loops?

Check the answer

They oscillate in antiphase.

Recall question 3

At an open tube end, what are the displacement and pressure conditions?

Check the answer

Displacement antinode and pressure node.

Try this next

Continue to the next lesson in this topic.

Two-source interference and Young double slit

Syllabus and review details

This lesson covers the listed H2 Physics 9478 outcomes. Topic 10 states no explicit exclusions. Topic 11(j) does not require knowledge of spectrometer structure or use. Graph interpretations distinguish time traces at one position from spatial profiles at one instant, and all small-angle equations are used only with their stated geometry.

  • GCE A-Level H2 PhysicsTopic 11(a) / Topic 11(b) / Topic 11(c) / Topic 11(d) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027