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  4. The frequency changes by 10 % as a sound source approaches a stationary observer with constant speed Vs. What would be percentage change in the frequency as the source recedes the observer with same speed (Vs < V) ?

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Q. The frequency changes by $10 \%$ as a sound source approaches a stationary observer with constant speed $V_{s}$. What would be percentage change in the frequency as the source recedes the observer with same speed $\left(V_{s} < V\right) ?$

Waves

Solution:

$f^{\prime}=\frac{v}{v-v_{s}} f_{0}$
$f$ is such that $f=\frac{110}{100} f_{0}$
When $\frac{v}{v-v_{s}}=\frac{110}{100}$
$100 v=110 v-110\, v_{s} $
$v=11 \,v_{s}$
When source is received
$\frac{v}{v+v_{s}} f_{0}=\frac{x}{100} f_{0}$
Putting $v=11 v_{s}$
$\frac{11}{12} \times 100=x$
$x=91.66 \% $
$\% $ change $=100-91.66=8.5 \%$