# A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first...

## Question:

A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork?

## Beat Frequency

Waves propagate according to the classical wave equation. This is a linear second-order partial differential equation with no inhomogeneities. It follows that linear combinations of independent solutions are also solutions. That is, it is possible to superpose waves. Physically this means that any number of waves can co-exist in the same region of space at the same time. As a result, waves cause interference in space and time. Interference in space will appear as interference fringes as in Young's double-slit experiment and that in time manifests as beats.

If 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 interfernce 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)} {/eq}.

Here there are two tuning forks, one with a frequency {eq}\displaystyle {\nu_1\ Hz} {/eq}.

It makes beats with another fork of frequency {eq}\displaystyle {\nu_2=256 Hz} {/eq}.

The beat frequency is {eq}\displaystyle {\nu=4\ Hz} {/eq}.

Therefore,

{eq}\displaystyle {\nu_1=252} {/eq} or {eq}\displaystyle {\nu_1=260} {/eq}.

To determine which is the correct frequency the first fork is loaded with wax. The increased inertia will lower its frequency. Then it is found that the beat frequency increases. This can only mean that the other fork has a higher frequency. Hence the unknown frequency of the second fork must be,

{eq}\displaystyle {\nu_2=252\ Hz} {/eq}. 