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A tuning fork of unknown frequency prod...

A tuning fork of unknown frequency produces 5 beats per second with another tuning fork. The second fork can cause a closed organ pipe of length 40 cm to vibrate in its fundamental mode. The beat frequency decreases when a small amount of wax is dropped on the first fork. Find out the frequency of the first tuning fork. Given, speed of sound in air ` = 320 m * s^(-1)` .

Text Solution

Verified by Experts

Length of the organ pipe = 40 cm = 0 . 4 m
So, its fundamental frequency ` = (V)/(4l) = (320)/(4 xx 0 . 4) = 200 ` Hz
`:.` The frequecny of the 2nd tuning fork,
`n_(2) = 200 ` Hz
Clearly, the frequency of the 1st tuning fork is
` n_(1) = (200 + 5) Hz or (200 - 5 )` Hz ,
i.e., ` n_(1) = 205 Hz or, 195` Hz
The wax would decrease the frequency of the 1st fork. This decreased frequency must be closer to `n_(2) = 200 ` Hz , as the beat frequency also decreases. So the frequency of the 1st tuning fork is `n_(1) = 205` Hz .
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