An engine sounding a whistle of frequency 1152 ~Hz is receding from a stationary observer at 72 ~km / hour. If velocity of sound in air is 340 ~m / s , then the frequency of note heard by the observer is

Step-by-step solution

Concept Source is moving away from a stationary observer f ^ = f ( v v + v _ s ) Where, f is the true frequency v is the speed of sound and v_s is the speed of source. Given, f =1152, v =340 ~m / s and v _ s = 72 10^3 3600 ~m / s =20 ~m / s the apparent frequency f ^ =1152 (340) (340+20) Hz =1088 ~Hz