A tuning fork of unknown frequency makes 5 beats per second with … - Vidyadip

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 40cm 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 320m/s.

✓

Answer

Given length of the closed organ pipe, $\text{l}=40\text{cm}=40\times10^{-2}\text{m}$

$\text{V}_\text{air}=320$

So, its frequency $\rho=\frac{\text{V}}{4\text{l}}=\frac{320}{4\times40\times10^{-2}}=200\ \text{Hertz}.$

As the tuning fork produces 5 beats with the closed pipe, its frequency must be 195Hz or 205Hz. Given that, as it is loaded its frequency decreases.

So, the frequency of tuning fork = 205Hz.

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