SAT Physics Subject Test

Chapter 13 Waves


Just as standing waves can be set up on a vibrating string, standing sound waves can be established within an enclosure. In the figure below, a vibrating source at one end of an air-filled tube produces sound waves that travel the length of the tube.

These waves reflect off the far end, and the superposition of the forward and reflected waves can produce a standing wave pattern if the length of the tube and the frequency of the waves are related in a certain way.

Notice that air molecules at the far end of the tube can’t oscillate horizontally because they’re up against a wall. So the far end of the tube is a displacement node. But the other end of the tube (where the vibrating source is located) is a displacement antinode. A standing wave with one antinode (A) and one node position (N) can be shown as follows:

Although sound waves in air are longitudinal, here we’ll show the wave as transverse so that it’s easier to determine the wavelength. Since the distance between an antinode and an adjacent node is always  of the wavelength, the length of the tube, L, in the figure above is  the wavelength. This is the longest standing wavelength that can fit in the tube, so it corresponds to the lowest standing wave frequency, the fundamental

The next higher-frequency standing wave that can be supported in this tube must have two antinodes and two nodes.

In this case, the length of the tube is equal to 3(λ′), so

Here’s the pattern: Standing sound waves can be established in a tube that’s closed at one end if the tube’s length is equal to an odd multiple of λ. The resonant wavelengths and frequencies are given by the equations

If the far end of the tube is not sealed, standing waves can still be established in the tube, because sound waves can be reflected from the open air. A closed end is a displacement node, but an open end is a displacement antinode. In this case, then, the standing waves will have two displacement antinodes (at the ends of the tube), and the resonant wavelengths and frequencies will be given by

Notice that, while an open-ended tube can support any harmonic, a closed-end tube can only support odd harmonics.

Questions 16-18

A closed-end tube resonates at a fundamental frequency of 343 Hz. The air in the tube is at a temperature of 20°C, and it conducts sound at a speed of 343 m/s.

16. What is the length of the tube?

17. What is the next higher harmonic frequency?

18. Answer the questions posed in questions 15 and 16 assuming that the tube was open at its far end.

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16.    For a closed-end tube, the harmonic frequencies obey the equation fn = . The fundamental corresponds to n = 1, so

17.    Since a closed-end tube can support only odd harmonics, the next higher harmonic frequency (the first overtone) is the third harmonic, f3, which is 3f1 = 3(343 Hz) = 1,029 Hz.

18.    For an opened-end tube, the harmonic frequencies obey the equation fn = nv/(2L). The fundamental corresponds to n = 1, so

Since an opened-end tube can support any harmonic, the first overtone would be the second harmonic, f2 = 2f1 = 2(343 Hz) = 686 Hz.