A stationary source emits sound of frequncy f_(0) = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms^(-1). The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound in air is 330 ms^(-1) and the car reflects the sound at the frequency it has received.) {:(0,1,2,3,4,5,6,7,8,9),(,,,,,,,,,):}

A stationary source emits sound of frequncy f_(0) = 492 Hz. The sound

1.	A stationary source emits sound of frequncy f_(0) = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms^(-1). The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound in air is 330 ms^(-1) and the car reflects the sound at the frequency it has received.) {:(0,1,2,3,4,5,6,7,8,9),(,,,,,,,,,):}
Answer» Solution :For reflected sound we can assume it to be a case of source and observer approaching towards each other. If U is speed of car and `UPSILON` is speed of sound then FREQUENCY received after reflection can be WRITTEN as follows : `f = f_(0) ((upsilon + u)/(v - u))` Beat frequency can be written as follows : `Delta f = f_(0) ((upsilon + u)/(upsilon - u)) - f_(0)` `rArr""Delta f = f_(0) (2u)/(upsilon - u)` Substituting values we get the following : `rArr""Delta f = 492 xx ((2 xx 2)/(330 - 2)) = 6`