# The tuning forks A and B gives 3 beats per second. The frequency of A is 512 Hz. The second fork...

## Question:

The tuning forks A and B gives 3 beats per second. The frequency of A is 512 Hz. The second fork B is filed so that its beat frequency reduces to 2 beats per second. What is the frequency of B?

When two wave sources with slightly differing frequencies {eq}\displaystyle {\nu_1} {/eq} and {eq}\displaystyle {\nu_2} {/eq} generate waves simultaneously and these waves are superposed then an interference effect in time will occur. The intensity is found to oscillate with time with a frequency {eq}\displaystyle {\nu} {/eq} called the beat frequency. It is given by,

{eq}\displaystyle {\nu = \pm (\nu_1-\nu_2)}-----------(1) {/eq}.

The frequency of the fork A is given to be {eq}\displaystyle {\nu_A=512\ Hz} {/eq}. It gives 3 beats with the fork B. Therefore from (1) it follows that either {eq}\displaystyle {\nu_B=515\ Hz} {/eq} or {eq}\displaystyle {\nu_B=509\ Hz} {/eq}.

Now B is filed so that its inertia reduces. Then it will tend to oscillate more rapidly. That is the filing increases the frequency of B. Now it is seen that the beat frequency reduces to 2 Hz. Thus the filing has caused a convergence between the two frequencies. This can only mean that B has a lower frequency initially. Hence we have,

{eq}\displaystyle {\nu_B=509\ Hz} {/eq}. 