Question

A tuning fork of unknown frequency makes 5 beats per second with another tuning fork which can cause a closed organ pipe of length 40 cm to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded with wax. Find its original frequency. The speed of sound in air is 320 m/s.

Answer 1

Given length of the closed organ pipe,

I=40 cm $=40\times {10}^{-2}m$

$V_{air}=320$

So its frequency,

$n=\frac{V}{4I}=\frac{320}{4\times 40\times {10}^{-2}}=200Hz$

As the tuning fork produces 5 beats with the closed pipe, its frequency must be 195 Hz or 205 Hz.

Given that,as it is loaded its frequency decreases, so the frequency of tuning fork

=205 Hz.