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15.3Standing waves in pipes

The standing waves discussed so far in this chapter apply to string musical instruments such as guitars and violins. But standing waves can also be produced within pipes which can have open or closed ends. These standing waves apply to wind instruments such as flutes and clarinets.

Consider a pipe of length L that is open at both ends; that is, the end conditions are antinode–antinode (Figure 15.6). A flute is an example of this. A travelling wave sent down the pipe will reflect from the ends (even though they are open—you may want to look further into this), and we again have the condition for the formation of a standing wave, namely two waves travelling in opposite directions.

Figure 15.6: a A pipe with both ends open has two antinodes at the open ends and a node in the middle. b The second harmonic in an open pipe.

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The top part of Figure 15.6a represents the first harmonic in a pipe with open ends. The dots represent molecules of air in the pipe. The double-headed arrows show how far these molecules oscillate back and forth (the amplitude of the oscillations). We see that the molecules at the ends oscillate the most: they are at antinodes. The molecules in the middle of the pipe do not oscillate at all: they are at a node. We have antinodes at the open ends, and there is a node in the middle. The lower diagram is how we normally represent the standing wave in the pipe—you must understand that it represents what the top diagram shows. Figure 15.6b represents the second harmonic.

Exam tip

You must be able to explain how molecules move in a longitudinal standing wave such as one in a pipe.

Note that these diagrams also give the harmonics for the case of a string with both ends free. (Imagine a vertical string.)

The case of a pipe with both ends closed (which is not very useful) is similar to that of a string with ends fixed: Figure 15.7 shows the first and second harmonics.

The wavelength for pipes with both ends closed or both ends open is:

λ n=2 L n,n=1,2,3,4,…

Figure 15.7: The first and second harmonics for a pipe with both ends closed.

We consider finally the case of a pipe with one closed and one open end (in other words, end conditions node–antinode). This could apply to clarinets and some organ pipes. The closed end will be a node and the open end an antinode.

Figure 15.8: The first two harmonics in a pipe with one open and one closed end.

Figure 15.8 shows the first two harmonics. The distance between a node and an antinode is a quarter of a wavelength, and so the wavelength of the first harmonic is given by:

1×λ 1 4=L⇒λ 1=4 L=4 L 1

The wavelength of the next harmonic is:

3×λ 3 4=L⇒λ 3=4 L 3

Notice that only ‘odd’ harmonics are present. In general, the allowed wavelengths are:

λ n=4 L n,n=1,3,5,…

This formula also gives the wavelength for the case of a string with one fixed and one free end.

Table 15.1 summarises the relationships for standing waves in strings and pipes.

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String of length L | Both ends fixed or both free: λ n=2 L n n=1,2,3,4,… One end fixed, the other free: λ n=4 L n n=1,3,5,…
Pipe of length L | Both ends open or both closed: λ n=2 L n n=1,2,3,4,… One end closed, the other open: λ n=4 L n n=1,3,5,…

Table 15.1: Wavelengths for allowed harmonics for standing waves in strings and pipes.

CHECK YOURSELF 4

In a pipe with one end closed and the other open, a particular harmonic has frequency 126 Hz. The next harmonic has frequency 210 Hz. What is the frequency of the first harmonic in this pipe?

Worked Example 15.3

A pipe X has one open and one closed end. Another pipe Y has both ends open. The first harmonic in both pipes has the same wavelength. Determine the ratio of the length of X to the length of Y.

Answer

In X the wavelength is λ = 4L x and in Y, λ = 2L Y. Hence:

4 L x=2 L y⇒L x L y=1 2.

Worked Example 15.4

A source of sound of frequency 2100 Hz is placed at the open end of a tube. The other end of the tube is closed. Powder is sprinkled inside the tube. When the source is turned on it is observed that the powder collects in heaps a distance of 8.0 cm apart.

aExplain this observation.

bUse this information to estimate the speed of sound.

Answer

aA standing wave is established inside the tube since the travelling waves from the source superpose with the reflected waves from the closed end. At the antinodes air oscillates the most and pushes the powder right and left. The powder collects at the nodes where the air does not move.

bThe heaps collect at the nodes, and the distance between nodes is half a wavelength. So the wavelength is 16 cm. The speed of sound is then:

v=f λ=2100×0.16=336≈340 m s−1

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Worked Example 15.5

A tube with both ends open is placed inside a container of water, Figure 15.9. When a tuning fork above the tube is sounded, a loud sound comes out of the tube. The shortest length of the column of air for which this happens is L. The frequency of the tuning fork is 486 Hz, and the speed of sound is 340 m s−1.

aDetermine the length L.

bPredict the least distance by which the tube must be raised for another loud sound to be heard from the tube when the same tuning fork is sounding.

Figure 15.9: For worked example 15.5.

Answer

aSince the air column length L is the least possible for a standing wave to form, the standing wave must be in its first harmonic. The water surface makes the lower end of the tube closed so the wavelength is 4L. The wavelength is given by:

λ=340 486=0.6996≈0.700 m

and so:

L=0.6996 4=0.1749≈0.175 m

bThe length of the air column in the tube must be increased so that the next harmonic can fit. The wavelength must be the same as before (since the tuning fork is the same). In Figure 15.10 we show the original standing wave in blue and the new one in red. It is clear from the diagram that the extra length needed is half a wavelength, that is, 0.35 m.

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Figure 15.10: For worked example 15.5 part b.

Alternatively, you can argue that since the wavelength must be the same as before, the new air column length for the next harmonic is given by:

4 L′3=0.700 m

So L′ = 0.525 m.

The tube must be raised by 0.525 − 0.175 = 0.35 m.

CHECK YOURSELF 5

The top end of a vertical string is attached to an oscillator. The lower end is free. The speed of a transverse wave on the string is 24 m s−1. When the oscillator is set to a frequency of 5.0 Hz the string oscillates in the first harmonic mode. What is the length of the string?

NATURE OF SCIENCE

Physics is universal

The universality of physics is evident almost everywhere including in the theory of standing waves. From the time of Pythagoras onwards philosophers and scientists have used mathematics to model the formation of standing waves on strings and in pipes. The theory that we have developed here applies to simple vibrating strings and air columns, but it can be used to give detailed accounts of the formation of musical sound in instruments as well as the stability of buildings shaken by earthquakes.

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TEST YOUR UNDERSTANDING

12A pipe has one end closed and the other open. A particular harmonic has frequency 300 Hz. The next harmonic has frequency 420 Hz. What is the frequency of the first harmonic?

13Draw the standing wave representing the third harmonic standing wave in a tube with one closed and one open end.

14A glass tube is closed at one end. The air column it contains has a length that can be varied between 0.50 m and 1.20 m. A tuning fork of frequency 425 Hz is sounded at the top of the tube. Predict the lengths of the air column at which loud sounds would be heard from the tube. (Take the speed of sound to be 340 m s−1.)

15A glass tube with one end open and the other end closed is used in an experiment to determine the speed of sound. A tuning fork of frequency 427 Hz is used, and a loud sound is heard when the air column has length equal to 20.0 cm.

aCalculate the speed of sound.

bPredict the next length of air column when a loud sound will again be heard.

16Figure 15.11 shows a standing wave in a pipe. The red curve represents the standing wave at t = 0. The blue curve represents the standing wave at time t = T/10, where T is the period. What is the direction of the velocity at t = T/10 of a molecule of air whose equilibrium position is at:

apoint P

bpoint Q?

Figure 15.11: For question 16.

17A tube has both ends open. When a tuning fork is sounded next to one of the ends of the pipe, a loud sound comes out of the tube. The shortest length of the tube for which this happens is L. The frequency of the tuning fork is 586 Hz, and the speed of sound is 340 m s−1.

aDetermine the length L.

bPredict the least distance by which the tube must be lengthened for another loud sound to be heard from the tube when the same tuning fork is sounding.

18A pipe with both ends open has two consecutive harmonics of frequencies 300 Hz and 360 Hz. Take the speed of sound to be 340 m s−1.

aSuggest which harmonics are excited in the pipe.

bDetermine the length of the pipe.

19A pipe X has both ends open and a pipe Y has one open and one closed end. The second harmonic in X has the same frequency as the first harmonic in Y. Calculate the ratio of the length of pipe X to that of pipe Y.

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20Consider a string with both ends fixed. A standing wave in the second harmonic is established on the string, as shown in Figure 15.12. The speed of the wave is 180 m s−1.

Figure 15.12: For question 20.

aExplain the meaning of wave speed in the context of standing waves.

bConsider the vibrations of two points on the string, P and Q. The displacement of point P is given by the equation y = 5.0 cos (45πt), where y is in millimetres and t is in seconds. Calculate the length of the string.

cState the phase difference between the oscillation of point P and that of point Q. Hence write down the equation giving the displacement of point Q.

21A horizontal aluminium rod of length 1.2 m is hit sharply with a hammer. The hammer rebounds from the rod 0.37 ms later.

aExplain why the hammer rebounds.

bCalculate the speed of sound in aluminium.

cThe hammer created a longitudinal standing wave in the rod. Estimate the frequency of the sound wave by assuming that the rod vibrates in the first harmonic.