A tuning fork arrangement (pair) produces 4 beats/s with one fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is:
A
286 Hz
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B
292 Hz
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C
294 Hz
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D
288 Hz
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Solution
The correct option is B 292 Hz Let the frequency of the unknown fork be v.
As It produces, 4 beats with 288Hz,
⇒|v−288|=4
⇒v−288=±4⇒v=292 or 284Hz
waxing of fork decreases its frequency.
Since, beat frequency also drops, the frequency of unknown fork has to be, v=292Hz.