Question

A tuning fork of unknown frequency makes 3.00 beats per second with a standard fork of frequency 384Hz. The beat frequency decreases when a small piece of wax is put on a prong of the first fork. What is the frequency of this fork?

Short answer

The frequency of this fork is, 387Hz.

Step-by-step solution

The given data

1. The frequency of one fork is, f₁ = 384Hz.
2. Beat frequency is, f_beat = 3Hz.

Understanding the concept of the frequency

We can predict the frequency of the unknown fork from the frequency of standard form using the formula for beat frequency. Then, using the relation between frequency and mass, we can conclude about this unknown frequency.

Formula:

The beat frequency of a wave,

${\mathit{f}}_{\mathbf{b}\mathbf{e}\mathbf{a}\mathbf{t}}\mathbf{=}{\mathit{f}}_{\mathbf{2}}\mathbf{-}{\mathit{f}}_{\mathbf{1}}$ …(i)

Calculation of the frequency of the fork

Since, $f\propto 1/\sqrt{m}$, the wax will decrease the frequency of the fork, so the decrease in frequency brings it closer to 384Hz, and therefore, its original frequency must have been bigger than 384Hz, so using equation (i), the value of other fork is given as:

$f_2=f_{beat}+f_1$

Substitute all the value in the above equation.

$\begin{array}{rcl}{f}_2& =& f_{beat}+f_1\\ & =& 3Hz+384Hz\\ f_2& =& 387Hz\end{array}$

Therefore, we can say that, when mass is added to this fork, its frequency decreases. Since the beat frequency also decreases, the frequency of the first fork must be greater than the frequency of the second fork. So, it must be 387Hz.