Stationary Waves A Level: Nodes, Antinodes and Harmonics

What is a stationary wave?

A stationary wave (also called a standing wave) is a wave pattern that does not travel along. The peaks and troughs stay in fixed places, each point on the wave just oscillates up and down, and the amplitude of that oscillation depends on where you are. Guitar strings, organ pipes and microwave ovens all rely on them.

The key idea is simple. A progressive wave carries energy from one place to another. Send one down a string and it reflects from a fixed end, so a wave comes back the other way. At every instant the displacement of the string is the sum of the two. That is the principle of superposition. When the frequency and the length of the string fit together just right, the sum is a steady pattern with points that never move and points that move a lot.

This guide covers how the pattern forms, nodes and antinodes, phase, harmonics on strings and in pipes, and the resonance tube experiment. The Module 4 hub lists the related topics.

How a stationary wave forms: superposition

Take two progressive waves with the same frequency, wavelength and amplitude, travelling in opposite directions along the same line. Superposition says the resultant displacement at each point is the sum of the two displacements, remembering that displacements can be positive or negative.

Where the two waves arrive in phase, the displacements add and you get constructive interference: a large oscillation. Where they arrive in antiphase, the displacements cancel and you get destructive interference: no movement at all. Because the waves are travelling in opposite directions, the points where this happens stay put. That is what makes the pattern stationary.

A quarter of a period after t = 0 each wave has moved λ/4, one to the right and one to the left, so they are opposite in sign at every point and the resultant is zero everywhere. The nodes, where the resultant is always zero, never move.

In practice the second wave nearly always comes from reflection: from the fixed end of a string, or from the closed or open end of a pipe. Some energy is lost at each reflection, so the two waves are never perfectly equal and the nodes of a real stationary wave have a small, non-zero amplitude.

Nodes, antinodes and phase

A node is a point on a stationary wave where the amplitude is zero: the two waves always cancel there. An antinode is a point where the amplitude is a maximum: the two waves always reinforce each other there. Between them, the amplitude changes smoothly.

• The distance between two adjacent nodes is λ/2. The distance between two adjacent antinodes is also λ/2.
• The distance from a node to the next antinode is λ/4.
• All points except the nodes oscillate at the frequency of the two progressive waves.
• The amplitude is not the same at every point, and it is largest at the antinodes. This is the main difference from a progressive wave.

The phase rules are a favourite exam topic, and they are easy to get right if you picture the loops. A loop is the region between two adjacent nodes.

On a progressive wave the phase changes steadily, by 2π over one wavelength. On a stationary wave it stays the same within a loop and flips by π at each node.

Harmonics on a string fixed at both ends

When a string is fixed at both ends, those ends cannot move, so each end must be a node. Only certain stationary waves fit. These are called the harmonics, and the lowest-frequency one is the fundamental (or first harmonic).

For the nth harmonic there are n loops, so the length of the string is L = nλ/2, which gives λ = 2L/n. Since the wave speed on the string is v = fλ, the frequencies are:

The speed of waves on a stretched string depends on its tension T and its mass per unit length μ, and is given by v = √(T/μ). The fundamental frequency of a string is therefore f₁ = (1/2L)√(T/μ). Check the data and formulae booklet to see which of these expressions are printed, but make sure you can use them. A shorter, tighter or lighter string gives a higher fundamental frequency.

• Halve the length and the fundamental doubles (f ∝ 1/L).
• Quadruple the tension and the fundamental doubles (f ∝ √T).
• Use a string with four times the mass per unit length and the fundamental halves (f ∝ 1/√μ).

Harmonics in air columns: closed and open pipes

Sound waves in a pipe are longitudinal, but we draw them as if they were transverse to show the amplitude of the air displacement. A node is a place where the air does not move. An antinode is a place where the air oscillates with the largest amplitude.

The boundary conditions are different from a string. At a closed end the air cannot move, so there is a displacement node. At an open end the air is free to move, so there is a displacement antinode (very slightly beyond the end of the pipe, which we come back to below).

Closed at one end. The shortest pattern that fits has a node at one end and an antinode at the other, which is a quarter of a wavelength. So L = λ/4 for the fundamental. The next pattern adds half a wavelength, giving L = 3λ/4, then 5λ/4, and so on. Only odd multiples of the fundamental occur.

Open at both ends. There is an antinode at each end. The shortest pattern has an antinode, a node and an antinode, which is half a wavelength. So L = λ/2 for the fundamental, and every whole number of half wavelengths fits after that.

For the same length, the open pipe has a fundamental twice as high as the closed pipe, because its fundamental wavelength is 2L rather than 4L.

End correction. The displacement antinode at an open end is not exactly at the end of the pipe: it sits a short distance e beyond it, because the air just outside the end is also involved in the oscillation. The effective length of a pipe closed at one end is therefore L + e, so the fundamental satisfies L + e = λ/4. Exam questions often tell you to ignore it, but it matters in the resonance tube experiment.

Practical: measuring the speed of sound with a resonance tube

This classic experiment can be practised in the lab and tested in the written papers. A vertical tube is partly filled with water, so the water surface acts as the closed end. A tuning fork of known frequency is held above the open end. You change the length of the air column and listen for resonance, a loud note that occurs when the column length fits a stationary wave.

1. Stand a tall glass tube in a cylinder of water so you can change the length L of the air column.
2. Strike a tuning fork of known frequency f and hold it just above the open end.
3. Slowly raise the tube (or lower the water level) from a short air column and listen for the first loud note. Measure the length L₁ from the open end to the water surface.
4. Continue to the next loud note and measure L₂. Repeat the whole procedure several times and take means.
5. Calculate λ = 2(L₂ − L₁) and then v = fλ. Repeat with forks of different frequencies if you can.

Why use the difference between two lengths? Because of the end correction. At the first resonance, L₁ + e = λ/4. At the second, L₂ + e = 3λ/4. Subtracting, L₂ − L₁ = λ/2, and the unknown e has cancelled.

An alternative method uses several forks and plots a graph. For the first resonance, L + e = v/4f, so L = (v/4)(1/f) − e. A graph of L against 1/f is a straight line with gradient v/4 and a negative intercept on the L axis equal to −e. This uses all your data and gives the end correction directly.

Progressive and stationary waves compared

Comparing the two types of wave is a standard question.

The energy point deserves a careful sentence. There is no net transfer of energy along a stationary wave, because the two progressive waves carry equal amounts of energy in opposite directions. The energy is still there, swapping between kinetic and potential forms.

Worked examples

Common mistakes

• Using node spacing as the wavelength. Adjacent nodes are λ/2 apart, not λ. This is the most common slip in calculations involving microwaves, strings and sound.
• Forgetting that a closed pipe has only odd harmonics. The second resonance of a closed pipe is the third harmonic, at three times the fundamental, not twice.
• Mixing up the formulae. f = nv/2L is for a string fixed at both ends and a pipe open at both ends. The closed pipe uses 4L.
• Getting the phase of adjacent loops wrong. It is π rad (antiphase), not 90° or 2π.
• Saying there is no energy in a stationary wave. There is no net transfer of energy along it. The energy is still stored in the oscillation.
• Treating the picture of a pipe as a transverse wave. The curves show the amplitude of longitudinal displacement.

How OCR A examines stationary waves

Stationary waves sit in Module 4 (Electrons, waves and photons). Expect these question styles.

• Explanation questions. Explain how a stationary wave forms, and how it differs from a progressive wave.
• Diagram questions. Draw the first, second or third harmonic on a string, or the displacement pattern in a closed or open pipe, and label nodes and antinodes.
• Calculation questions. Use f = nv/2L, v = fλ and v = √(T/μ), or work out wavelength from node spacing using microwaves, sound or a string.
• Data and practical questions. Analyse a graph of L against 1/f from a resonance tube, and comment on uncertainty.

Practical work with stretched strings, resonance tubes and microwaves is the usual way to meet this topic in the lab, and the written papers test the methods, analysis and evaluation of that work through your OCR A practical activity groups (PAGs). Be ready to describe how you would measure the speed of sound and improve its precision.

There are synoptic links. Stationary waves connect to simple harmonic motion, because every point on the wave is a simple harmonic oscillator, and to ultrasound and acoustic impedance through reflection at boundaries.

Practice questions

1. A microwave transmitter and a reflecting metal plate produce a stationary wave. The distance between adjacent nodes is 6.1 cm. Calculate the frequency of the microwaves. (c = 3.00 × 10⁸ m s⁻¹)3 marks

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   Answer: 2.46 × 10⁹ Hz

   λ = 2 × 0.061 = 0.122 m. f = c/λ = 3.00 × 10⁸ / 0.122 = 2.46 × 10⁹ Hz.

2. A string of length 0.80 m has a mass per unit length of 1.0 × 10⁻³ kg m⁻¹ and is under a tension of 64 N. Calculate the speed of waves on the string and the frequency of its fundamental mode.4 marks

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   Answer: v = 253 m s⁻¹ and f₁ = 158 Hz

   v = √(T/μ) = √(64 / 1.0 × 10⁻³) = √64000 = 253 m s⁻¹. Fundamental: λ = 2L = 1.60 m, so f₁ = v/λ = 253 / 1.60 = 158 Hz.

3. A pipe of length 0.30 m is closed at one end. Take the speed of sound as 343 m s⁻¹ and ignore end corrections. Calculate the frequencies of the fundamental and the next harmonic that can resonate.3 marks

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   Answer: 286 Hz and 858 Hz

   f₁ = v/4L = 343 / (4 × 0.30) = 286 Hz. The next is the third harmonic: 3 × 285.8 = 858 Hz.

4. In a resonance tube experiment, a student plots the length L of the air column at first resonance against 1/f for several tuning forks. The line has a gradient of 85 m s⁻¹ and an intercept on the L axis of −0.012 m. Calculate the speed of sound and the end correction.3 marks

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   Answer: v = 340 m s⁻¹ and e = 0.012 m (1.2 cm)

   L + e = v/(4f), so L = (v/4)(1/f) − e. Gradient = v/4, so v = 4 × 85 = 340 m s⁻¹. The intercept is −e, so e = 0.012 m.

5. Describe how the amplitude and phase of the oscillation of points on a stationary wave vary along the wave, and state two differences between a stationary wave and a progressive wave.5 marks

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   Answer: Amplitude: zero at nodes, maximum at antinodes, varying in between. Phase: points within a loop are in phase; points in adjacent loops are in antiphase (π rad). Differences: a stationary wave has no net energy transfer along it, while a progressive wave does; a stationary wave has nodes and varying amplitude, while a progressive wave has the same amplitude everywhere.

   Another acceptable difference: phase changes continuously along a progressive wave but not within a loop.

Next steps

Once you are comfortable with nodes, antinodes and the harmonic formulae, test yourself with timed questions. Stationary waves link to simple harmonic motion, and they share the idea of reflection with ultrasound and acoustic impedance. For the rest of the topic list, see the Module 4 hub. When you are ready, practise this topic with step-by-step marking. It is free to start.

Related guides

Module 5 · OscillationsSimple Harmonic Motion Equations: Graphs, Energy and Worked Examples

Module 6 · Medical imagingAcoustic Impedance and Ultrasound: A-Level Physics Guide

Module 4 · Quantum physicsPhotoelectric Effect A Level: Equation, Graph and Questions (OCR A)