Question

A railway engine and a car are moving parallel but in opposite direction with velocities 144 km/hr and 72 km/hr respectively. The frequency of engine’s whistle is 500 Hz and the velocity of sound is 340 m s^-1 . Calculate the frequency of sound heard in the car when(i) the car and engine are approaching each other(ii) both are moving away from each other.

Answer 1

This change of frequency is called the Doppler effect.

   v = velocity of sound

Apparent frequency  f' = f₀  (v +- v_observer) / (v -+ v_source)
             
1)  f ' = 500 * (340 + 144*5/18) / (340 - 72*5/18) Hz
         = 500 * 380/320  Hz

2)  f ' = 500 (340 - 144*5/18) / (340 + 72*5/18)
         = 500 * 300/360  Hz

Answer 2

vl = 72x5/18 = 20m/s

vs = 144x5/18 = 40m/s

v = 340m/s

f = 500Hz

Case 1:

Car and engine are apporching towards each other.......

therefore, v' = f [v - vl/v - vs]

=> v' = 500[340 + 20 / 340 - 40]

=> v' = 500 [360/300]

=> v' = 600 Hz

Case 2:

Car and engine are moving away from each other......  

here as listner is moving away from the source of sound

so vl = -20 and vs = -40

v' = 500 [340+(-20)/ 340 -(-40)]

v' = 500 [320/380]

v' = 421Hz

Hope u understand...