Each of two persons has a whistle of frequency 500 Hz . One person is at rest at a particular place nad the second person recedes from him with a velocity of ` 1.8 m * s^(-1)`
If both of them blow whistles , how many beats will be heard by each of them ? Velocity of sound = 330 m ` s^(-1)`

Explanation

Both of them listen to the sound of frequency 500 Hz form their own whistles .
In the first case , let us suppose that the first person is listenting to the sound coming from the whistle of the second person .
Here the velocity of the listener , ` u_(o) = 0 `
Since the source is receding ,
the velocity of the source ,
` u_(s) = - 1.8 * s^(-1)`
` therefore ` Apparent frequency ,
` n. = (V)/(V - u_(s)) xxn = (330)/(330 - (-1.8)) xx 500 = 497.29` Hz
` therefore ` Number of beats per second ` = n- n. = 500 - 497.29`
` = 2. 71 ~~3 `
In the second case , let us suppose that the second person is listening to the sound coming from the whistle of the first person .
Here the velocity of the source , ` u_(s) = 0 `
Since the listener is receding ,
The velocity of the listener , ` u_(0) = - 1.8 * s^(-1)`
` therefore ` Apparent frequency ,
` n. = ( V + u_(o))/(V) xxn = (330 - 1.80/(330 ) xx 500 = 497.27 Hz`
` therefore ` Number of beats per second = n - n. = 500 - 497.27
` = 2.73 ~~ 3`