Given, velocity of source of sound =50ms−1 ; frequency measured by observer = 500 Hz Now, frequency of sound when source is moving towards stationary observer is f′=f(
v
v−vs
) → (1) Frequency of sound when source is moving away from the stationary observer is f′′=f(
v
v+vs
) → (2) Using Eqs. (1) and (2), we have
f′
f′′
=
f(
v
v−vs
)
f(
v
v+vs
)
⇒
f′
f′′
=
(v+vs)
(v−vs)
Substitute v=350ms−1;vs=50ms− f′=500Hz
500
f′′
=
350+50
350−50
⇒
500
f′′
=
400
300
⇒f′′=
300×500
400
Therefore, f′′=375Hz Thus, apparent frequency of sound as heard by the observer when source is moving away from him is 375 Hz.