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Question

Figure shows an air filled, acoustic interfero meter, used to demonstrate the interference of sound waves. Sound source S is an oscillating diaphragm; D is a sound detector, such as the ear or a microphone. Path SBD can be varied in length, but path SAD is fixed. At D, the sound wave coming along path SBD interferes with that coming along path SAD. In one demonstration, the sound intensity at D has a minimum value of 100 units at one position of the movable arm and continuously climbs to a maximum value of 900 units when that arm is shifted by 1.65cm. Find the frequency of the sound emitted by the source.
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Solution

When the right side of the instrument is pulled out a distance d, the path length for sound waves increases by 2d.
since the interference pattern changes from a minimum to the next maximum, this distance must be half a wavelength of the sound.
So 2d=λ/2 where λ is the wavelength.
Thus λ=4d and, if v is the speed of sound, the frequency is
f=v/λ=v/4d=(343m/s)/4(0.0165m)=5.2×103Hz

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