9478 / 2027
Superposition overview

Lesson 2 of 9

Build a standing wave

  • I can explain standing-wave formation graphically, identify nodes and antinodes, and distinguish sound pressure and displacement patterns.Syllabus

    Syllabus 9478, 11(c). Explain standing-wave formation graphically, identify nodes and antinodes, and distinguish sound pressure and displacement patterns.

Equal-amplitude waves of the same frequency travelling in opposite directions can form a standing wave. Its fixed pattern of oscillation amplitude differs from the travelling profiles that produce it.

Take two waves in the same region and oscillation direction, with the same wavelength and speed but opposite propagation directions. At every position and instant, add their signed displacements. The result is also called a stationary wave.

Add the profiles through a cycle

Each component in this supplied model has amplitude 2.0 mm, wavelength 0.80 m and period T = 0.20 s. The view covers x = 0 to 0.80 m.

Two travelling profiles form one standing pattern

Component 1 travels right; component 2 travels left. Each has amplitude 2.0 mm. Brown long dashes and green short dashes distinguish the components; solid blue is their point-by-point sum. Every panel shares the same axes.

t = 0

Standing-wave construction at t = 0The horizontal axis is position from zero to 0.80 metres; the vertical axis is signed displacement in millimetres, from negative four to positive four. Components each have amplitude two millimetres and travel oppositely. The two component profiles coincide and are labelled together. The summed amplitude at an antinode is twice the component amplitude. At half a period all displacements reverse from the initial profile. Node positions are fixed in every panel. No envelope is being substituted for the instantaneous sum.0.000.200.400.600.80-4-20+2+4Displacement / mmPosition x / m

Component 1 = component 2 in this snapshot, so their dashed profiles coincide. The blue resultant is twice either component.

t = T/4 = 0.050 s

Standing-wave construction at t = T/4 = 0.050 sThe horizontal axis is position from zero to 0.80 metres; the vertical axis is signed displacement in millimetres, from negative four to positive four. Components each have amplitude two millimetres and travel oppositely. At one quarter period they are equal and opposite, so the whole blue sum is instantaneously zero. Only the three marked positions zero, 0.40 and 0.80 metres are permanent nodes. The antinodes at 0.20 and 0.60 metres are moving downward and upward respectively, as the direction arrows show. Node positions are fixed in every panel. No envelope is being substituted for the instantaneous sum.0.000.200.400.600.80-4-20+2+4Displacement / mmPosition x / m

The small arrows show particle-motion directions at the antinodes. An instantaneous zero displacement everywhere does not make every point a node.

t = T/2 = 0.100 s

Standing-wave construction at t = T/2 = 0.100 sThe horizontal axis is position from zero to 0.80 metres; the vertical axis is signed displacement in millimetres, from negative four to positive four. Components each have amplitude two millimetres and travel oppositely. The two component profiles coincide and are labelled together. The summed amplitude at an antinode is twice the component amplitude. At half a period all displacements reverse from the initial profile. Node positions are fixed in every panel. No envelope is being substituted for the instantaneous sum.0.000.200.400.600.80-4-20+2+4Displacement / mmPosition x / m

Component 1 = component 2 in this snapshot, so their dashed profiles coincide. The blue resultant is twice either component.

The two progressive profiles and their sum share the same scales at three times. At the first and third times the component curves coincide; at T/4 they are equal and opposite. The zero resultant at T/4 is an instantaneous state, not a new set of nodes.
  • At t = 0, the resultant displacements at x = 0, 0.20, 0.40, 0.60 and 0.80 m are 0, +4.0, 0, -4.0 and 0 mm.
  • At t = T/4 = 0.050 s, the whole resultant profile is at equilibrium. The two component profiles are not both zero; they cancel point by point.
  • At t = T/2 = 0.100 s, the nonzero resultant displacements have reversed sign.

Identify amplitude, nodes and antinodes

A node has zero displacement at all times and therefore zero oscillation amplitude. The fixed nodes here are x = 0, 0.40 and 0.80 m. An antinode has the greatest oscillation amplitude: here 4.0 mm at x = 0.20 and 0.60 m.

Amplitude describes the full oscillation, not the displacement in one frame. At T/4, an antinode passes through equilibrium at its greatest speed; it has not become a node.

Why does the zero profile not make every point a node?

A node remains at equilibrium throughout the cycle. Most points in the T/4 frame have nonzero displacements before and after it. Their zero displacement is temporary; only the fixed node positions have zero oscillation amplitude.

Adjacent nodes: separation = λ/2
Adjacent antinodes: separation = λ/2
A node to its neighbouring antinode: λ/4

Points within one loop oscillate in phase. Points in neighbouring loops oscillate in antiphase: when one loop moves up, the next moves down. The nodes separating them do not oscillate.

The ideal standing pattern has no net time-averaged energy transport along it. Energy is still present and exchanges locally. Real driven demonstrations need an energy input to replace losses; this does not turn the fixed node pattern into travelling crests.

The graphical model explains formation. The string and microwave experiments and air-column experiment show how the pattern can be observed and measured.

Put these in order

A string is fixed at both ends and driven at one of its natural frequencies. Put the explanation of how the standing wave forms in order.

Show the answer

1. A wave travels along the string and is reflected at the fixed end. The reflected wave has the same frequency, wavelength and speed as the incident wave, but travels the other way.

2. The incident and reflected waves travel in opposite directions through each other and superpose. At each point the two displacements add according to the principle of superposition.

3. Where the two waves always arrive in antiphase, the displacements cancel: a node. Halfway between two nodes the waves always arrive in phase.

4. Where they always arrive in phase, the displacements add to give the maximum amplitude: an antinode. Because the nodes and antinodes stay in the same places, the pattern does not travel.

5. The pattern stays in place, with adjacent nodes half a wavelength apart.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

State three differences between a stationary (standing) wave and a progressive wave. [3]

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Exam-style question

A string is fixed at one end and vibrated at the other. Explain how a stationary wave forms on the string. [3]

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Exam-style question

A graph of displacement against distance along a string shows a stationary wave at one instant. Adjacent nodes are 0.30 m apart. The vibrator frequency is 50 Hz. Find the speed of the progressive waves on the string. [3]

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