KCET 2016 Physics Solved Paper

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.