Three sources emitting sound waves of the same amplitude, have frequencies 400 Hz , 401 Hz and 402 Hz , respectively . Find out the number of beats heard per second .

Answer

` y_(1) = A sin (2 pi * 400 t) ; y_(2) = A sin (2 pi * 401 t)`
`y_(3) = A sin (2 pi * 402 t)`
`:. Y = y_(1) + y_(2) +y_(3)`
` = a [ sin ( 2 pi * 400t) + sin ( 2 pi * 401 t) + sin ( 2 pi * 402 t)]`
` = A [ sin (2 pi * 401 t) + 2 sin { 2 pi (402 + 400)/(2) t} * cos { 2pi (402 - 400)/(2) t }]`
` = a [ sin (2 pi * 401 t) + 2 sin ( 2pi * 401 t) cos 2 pi t]`
` = a (1 + 2 cos 2 pi t ) sin ( 2 pi * 401 t)`
Here, the amplitude of the resultant wave,
` A^(') = A (1 + cos 2 pi t) ` .
This is masimum when
` cos 2 pi t = 1 = cos 2 n pi [ n = 0 , 1 , 2 , 3 , .....]`
or, t = n = 0 , 1 , 2 , ....
Then , the time interval between two consecutive maxima is 1s; so the beat frequency = 1 per second .