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143. Problem 20.23 (RHK) In this problem we will study an acoustic interferometer.
It is a device used to demonstrate the interference of sound waves. S is a
source of sound (a loud speaker, for instance), and D is a sound
detector, such as the ear or a microphone. Path SBD can be varied in length, but
path SAD is fixed. The interferometer contains air, and it is found that the
sound intensity has a minimum value of
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at one
position of B and continuously climbs to a maximum value of
at
a second position 1.65 cm from the first. We have to find (a) the
frequency of the sound emitted from the source and (b) the relative
amplitudes of the wave arriving at the detector for each of the two positions of
B. (c) We have to explain how these waves have different amplitudes when
they originate at the same source.
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Solution: Click For PDF Version (a) Let the length of the path SAD be
where A and B are the amplitudes of the two waves that reach the detector D on travelling through different paths in the acoustic interferometer, and x and x + l are the path lengths. The detector D receives the resultant of the superposition of the two waves. The wave function of the resultant wave is
Let us assume that the variable path is at a position such that the resultant wave has a minimum. The minimum amplitude of the resultant wave will be proportional to
It is given that when the variable length is moved by 1.65 cm the resultant
wave at the detector has a maximum. The maximum amplitude of the resultant wave
will be proportional to
The additional phase difference will therefore be
This gives the wavelength
Speed of sound in air,
(b) It is given that the minimum value and the maximum value of the intensity are
This implies
From this we find
(c) The waves have different amplitudes as they travel through two different paths and have different path lengths.
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and the
length of the path SBD be
. Let the
waves that reach D from S following these two paths be represented
by the functions
. As
the intensity is proportional to modulus square of the amplitude,
. The
intensity of the maxima will be
, is
Therefore, the frequency of the wave is
