The frequency changes by 10 % as a sound source approaches a stat… - Vidyadip

MCQ

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 ........... $\%$

A
$10.5$
✓
$8.5$
C
$4.5$
D
$1.5$

✓

Answer

Correct option: B.

$8.5$

b
(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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