Question

what is beats show that the number of beats produced per second is equal to the difference in frequencies​

Answer 1

Beats :

When two interacting waves have slightly different frequencies the resultant distrubing at any point due to the superposition periodically fluctuates causing waxing and waning the resultant intensity.

The waxing and waning in the resultant intensity of two superposed waves of slightly different frequency are known as beats.

Let the displacement produced at a point by one wave be

${\sf{y_1 = A \: sin(2\pi f_1 t - \O_1)}}$

and the displacement produced at the point produced by the other wave of equal amplitude as

${\sf{y_2 = A \: sin(2\pi f_2 t - \O_2)}}$

By the principal of superposition, the resultant displacement is

${\sf{y = y_1 + y_2}}$

${\sf{y = A \: sin(2\pi f_1 t - \O_1) + A \: sin(2\pi f_2 t - \O_2)}}$

${\sf{y = 2A \: sin \bigg[\ 2\pi \bigg( \dfrac{f_1 + f_2}{2} \bigg)t - \bigg( \dfrac{\O_1 + \O_2}{2}\bigg)\bigg] cos2 \pi \bigg( \dfrac{f_1 - f_2}{2}\bigg) t}}$

${\sf{y = R\: sin \bigg[\ 2\pi \bigg( \dfrac{f_1 + f_2}{2} \bigg)t - \bigg( \dfrac{\O_1 + \O_2}{2}\bigg)\bigg] }}$

${\sf{ R = 2A \: cos \ 2\pi \bigg( \dfrac{f_1 + f_2}{2} \bigg)t }}$

The time for one beat is the time between consecutive maximum or minima

First maximum would occur when

${\sf{ cos \ 2\pi \bigg( \dfrac{f_1 + f_2}{2} \bigg)t = + 1}}$

${\sf{ Then \: \: \: \: 2\pi \bigg( \dfrac{f_1 + f_2}{2} \bigg)t = 0}}$

${ \sf{ \therefore t = 0}}$

For second maxima would occur when

${\sf{ cos \ 2\pi \bigg( \dfrac{f_1 + f_2}{2} \bigg)t = - 1}}$

${\sf{ Then \: \: \: \: 2\pi \bigg( \dfrac{f_1 + f_2}{2} \bigg)t = \pi}}$

${\sf{ \therefore t = \dfrac{1}{f_1 - f_2}}}$

${\sf{ The \: time \: for \: one \: beat \: = \bigg(\dfrac{1}{f_1 - f_2} - 0 \bigg) =\dfrac{1}{f_1 - f_2} }}$

Similarly it may also be shown that time between two consecutive minima is $\sf{\dfrac{1}{f_1 - f_2}}$

Hence, frequency of beat that is number of beats in one second or Beat frequency = f₁ - f₂