9702/May Jun/11/2015/Q29

A loudspeaker emitting sound of frequency f is placed at the open end of a pipe of length l which
is closed at the other end. A standing wave is set up in the pipe.

A series of pipes are then set up with either one or two loudspeakers of frequency f. The pairs of
loudspeakers vibrate in phase with each other.
Which pipe contains a standing wave?

Solution:

Answer: D.

For a stationary wave (resonance) to be formed in the tube, there should be a node (zero amplitude) at the closed end and an antinode (maximum amplitude) at the open end or at the loudspeaker (which is at an open end).

Let’s assume that the frequency f produces the fundamental mode of vibration. Since the same frequency is used in all cases, the wavelength will be the same.

We are told that when the frequency is f, a stationary wave is formed in the pipe of length l. For the fundamental mode, the wave formed is a quarter of a wavelength.

λ / 4 = L          giving wavelength λ = 4L

The wavelength will be the same in all of the cases.

Consider choice A:

For a stationary wave, there should be an antinode at the loudspeaker and an antinode at the open end of the pipe. This corresponds to half a wavelength.

For this case, λ / 2 = L             giving wavelength λ = 2L

BUT from above, we know that the wavelength = 4L while the length of the pipe is only l.

Thus, this is not possible.

Consider choice B:

This is similar to choice A as there should be an antinode at both ends. This is also not possible.

Consider choice C:

Here, a node is formed at the closed end at an antinode at the loudspeaker. This corresponds to a quarter of a wave.

For this case, λ / 4 = 2L                      giving wavelength λ = 8L

This does not correspond to the wavelength (= 4L) obtained initially. Hence, this is not correct.

Consider choice D:

Here, an antinode should be at both ends for a stationary wave to be formed. This corresponds to a half a wavelength.

For this case, λ / 2 = 2L                      giving wavelength λ = 4L

 
This is the only case where the wavelength corresponds to the original case.

Reference: PYQ - May/Jun 2015 Paper 11 Q29