When two tuning forks are stuck simultaneously, 6 beats per second are heard. The frequency of one fork is 560 Hz. A piece of wax is placed on the 560 Hz fork to lower its frequency slightly. If the beat frequency is increased, what is the correct frequency for the second fork

The correct frequenvy of the second fork is; ν2 = 566Hz

Explanation:

First of all, when two wave sources with slightly differing frequencies ν1 and ν2 generate waves at the same time and are superposed, then an interference effect will occur in time.

The intensity will be found to oscillate with time with a frequency (ν) called the beat frequency. It is depicted by:

ν = ± (ν1 −ν2).

In this question, there are two tuning forks, one with a frequency of let's say ν1=560Hz

The beat frequency is ν=6Hz

Therefore,

ν2=560 - 6 or ν2=560 + 6

i.e ν2=554 or ν2=566

For us to know the correct frequency, if the 560 Hz 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 unkown frequency of the second fork must be,

ν2 = 566Hz

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option number c is the right answer.