6.3 Stationary waves — Learn

Prerequisites: the principle of superposition and the wave equation from Section 6.1; the terms displacement, amplitude, period and frequency from Section 3.6.

By the end of this section you can: explain how a stationary wave forms; define a node and an antinode; find the node spacing; find the harmonics of a string fixed at both ends; use $v = \sqrt{\dfrac{T}{\mu}}$ for a wave on a string; and find the harmonics of an air column that is open or closed at one end.

1. How a stationary wave forms

Send a wave along a rope and let it reflect from the far end. The reflected wave travels back through the wave that is still arriving. The two waves have the same frequency and, if the reflection is good, about the same amplitude, but they travel in opposite directions.

Where they meet, the principle of superposition adds their displacements. The result is a stationary wave, also called a standing wave. It does not travel along the rope. Instead, some places stay still and others swing with the largest amplitude.

A node is a place where the displacement is always zero.
An antinode is a place where the amplitude is largest.

Neighbouring nodes are half a wavelength apart. A node and the next antinode are a quarter of a wavelength apart. So if you can measure the distance between two nodes, you can find the wavelength.

All the points on the stationary wave oscillate with the same frequency. Only the amplitude changes from place to place. Between two neighbouring nodes the points move together, and they are in phase. Across a node the motion flips over.

A stationary wave stores energy, but on average it transfers no energy along the rope. A progressive wave carries energy away; a stationary wave does not. This is the main difference between the two.

2. Waves on a string

A guitar string is fixed at both ends, so both ends must be nodes. The string can vibrate in several different shapes, called harmonics, and each shape must fit a whole number of half wavelengths between the ends:

L = n\frac{\lambda}{2} \qquad \lambda = \frac{2L}{n}

where $n = 1, 2, 3, \ldots$ The lowest frequency is the fundamental, or first harmonic, with $n = 1$ . It has one antinode in the middle. The second harmonic has two antinodes and twice the frequency. The frequency of the $n$ th harmonic is

f_n = \frac{nv}{2L}

The speed of a wave on a stretched string depends on the tension $T$ and the mass per unit length $\mu$ :

v = \sqrt{\frac{T}{\mu}}

Putting the last two equations together gives the frequency of the $n$ th harmonic:

f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}

A tighter string has a higher note, and a thicker string has a lower note. That is why the thick strings of a guitar sound the lowest notes.

Worked example 1. A guitar string has a tension of 81 N and a mass per unit length of $4.0 \times 10^{-4}\,\text{kg}\,\text{m}^{-1}$ . Find the speed of a wave on the string.

\begin{aligned} v &= \sqrt{\frac{T}{\mu}} = \sqrt{\frac{81}{4.0 \times 10^{-4}}} \\ &= \sqrt{2.025 \times 10^{5}} = 450\,\text{m}\,\text{s}^{-1} \end{aligned}

The wave travels along the string at $450\,\text{m}\,\text{s}^{-1}$ .

Worked example 2. The string in Worked example 1 is 0.650 m long and is fixed at both ends. Find the frequency of its fundamental note.

f_1 = \frac{v}{2L} = \frac{450}{2 \times 0.650} = 346\,\text{Hz}

The fundamental frequency is 346 Hz, to three significant figures.

Worked example 3. A string sounds a note of 220 Hz when its tension is 60 N. Find the tension needed for a note of 440 Hz.

The frequency is proportional to the square root of the tension, because $f \propto \sqrt{T}$ . So

\begin{aligned} \frac{f_2}{f_1} &= \sqrt{\frac{T_2}{T_1}} \\ T_2 &= T_1\left(\frac{f_2}{f_1}\right)^2 = 60 \times \left(\frac{440}{220}\right)^2 \\ &= 60 \times 4 = 240\,\text{N} \end{aligned}

To double the frequency, the tension must be four times larger.

3. Air columns

A wind instrument contains a column of air. A stationary wave forms in the air, and the note depends on the length of the column.

At an open end the air is free to move, so there is an antinode. At a closed end the air cannot move, so there is a node.

For a pipe open at both ends, both ends are antinodes. The distance between two neighbouring antinodes is half a wavelength, so

L = n\frac{\lambda}{2} \qquad f_n = \frac{nv}{2L}

with $n = 1, 2, 3, \ldots$ , exactly as for a string. All the harmonics are present.

For a pipe closed at one end, there is a node at the closed end and an antinode at the open end. The shortest fit is a quarter of a wavelength:

L = (2n - 1)\frac{\lambda}{4} \qquad f_n = \frac{(2n - 1)v}{4L}

with $n = 1, 2, 3, \ldots$ . This gives only the odd harmonics: the fundamental, the third, the fifth, and so on. The even harmonics are missing. A clarinet, closed at the mouthpiece end, sounds the odd harmonics, which is part of its characteristic tone.

Worked example 4. An organ pipe is open at both ends and is 0.800 m long. The speed of sound is $340\,\text{m}\,\text{s}^{-1}$ . Find the frequency of its fundamental note and of its second harmonic.

\begin{aligned} f_1 &= \frac{v}{2L} = \frac{340}{2 \times 0.800} = 213\,\text{Hz} \\ f_2 &= \frac{2v}{2L} = \frac{2 \times 340}{1.60} = 425\,\text{Hz} \end{aligned}

The fundamental is 213 Hz and the second harmonic is 425 Hz.

Worked example 5. A pipe of the same length is closed at one end. Find the frequency of its fundamental note and of its next possible harmonic.

\begin{aligned} f_1 &= \frac{v}{4L} = \frac{340}{4 \times 0.800} = 106\,\text{Hz} \\ f_3 &= \frac{3v}{4L} = \frac{3 \times 340}{3.20} = 319\,\text{Hz} \end{aligned}

The fundamental is 106 Hz and the next possible harmonic is the third, at 319 Hz. The second harmonic is missing, because it would need an antinode at the closed end, where the air cannot move.

Watch out. A stationary wave is not a wave travelling along the string or the pipe. The nodes never move, and no energy is carried from one end to the other. Do not use $v = f\lambda$ with the distance between a node and an antinode as if it were a wavelength. A node-to-antinode distance is a quarter of a wavelength, so multiply it by four to get $\lambda$ .

4. Try it: harmonics on a string

The figure draws the stationary wave on a string fixed at both ends. The sliders set the harmonic number, the length of the string and the phase of the motion. The wave speed is fixed at $450\,\text{m}\,\text{s}^{-1}$ , the speed on the string from Worked example 1.

Predict first. Do not move the sliders yet. Write down your answers.

If you change from the first harmonic to the second, what happens to the number of nodes and antinodes?
If you double the length of the string, what happens to the fundamental frequency?
Where are the nodes always found, whatever the harmonic?

Test. Set the length to 0.65 m and work through the first three harmonics. Read the wavelength and the frequency each time. Then drag the phase slider and watch which points stay still.

Explain. Why does a longer string give a lower note? Write two or three sentences. Use the idea that the string must fit a whole number of half wavelengths, so a longer string allows a longer wavelength.

Progress check

Return to the concert hall. The guitar string is 0.650 m long, its mass per unit length is $4.0 \times 10^{-4}\,\text{kg}\,\text{m}^{-1}$ , and its tension is 81 N. It is meant to sound the note E₄, 330 Hz.

Find the speed of a wave on the string.
Find the frequency of its fundamental note.
The measured frequency is higher than 330 Hz. Decide whether the technician should tighten or slacken the string.
Find the tension needed for exactly 330 Hz.

If question 1 goes wrong, use $v = \sqrt{\dfrac{T}{\mu}}$ . If question 2 goes wrong, use $f = \dfrac{v}{2L}$ . If question 4 goes wrong, use $T_2 = T_1\left(\dfrac{f_2}{f_1}\right)^2$ .

You should now be able to

explain how a stationary wave forms from two waves travelling in opposite directions;
define a node and an antinode, and give the distances between them;
explain how a stationary wave differs from a progressive wave;
find the harmonics of a string fixed at both ends;
use $v = \sqrt{\dfrac{T}{\mu}}$ for a wave on a string;
find the tension needed for a given frequency;
find the harmonics of an air column open at both ends;
find the odd harmonics of an air column closed at one end;
explain why the even harmonics are missing from a closed pipe.