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 $\left(V_s < V\right)$ is ........... $\%$
Diffcult
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(b)
$f^{\prime}=\frac{v}{v-v_s} f_0$
$f$ 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 $=100-91.66=8.5 \%$
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