Question 4
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?
Reference: Past Exam Paper – June 2015 Paper 11 Q29
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.
For option A, there is only 1 speaker thus 1 wave. The other end is open so the wave can't be reflected. A stationary wave can't form since this wave must interfere with another in the travelling in the opposite direction but this won't happen in the case of choice A.
ReplyDeleteIs my thinking correct?
in some way, you could say so. But that's not what is really important here.
DeleteWe need to understand that antinodes are at the open ends and nodes are closed ends and try to deduce the correct patterns possible.
Should there be 2 speakers on either end of an open tube to produce a stationary wave.
ReplyDeleteAll texts don’t show two speakers.
Kindly explain.
جزاك الله خيراً
i believe that yes, there should be 2. Unless we are clearly shown how reflection of the wave occurs which then result in the stationary wave.
DeleteEasy one
ReplyDeletethank you sir u r a life saver ����
ReplyDeleteIf the speakers are vibrating in phase, shouldn't the interference of the waves at midpoint of the tube result in zero phase difference i.e an antinode? but this will contradict the diagram given for choice D.
ReplyDeletethis in NOT a condition.
Deletethe conditions are:
there should be a node at the close end and
there should be an antinode at the open end