Question

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 $\left(V_s<V\right)?$

• A. $10.5\%$
• B. $8.5\%$
• C. $4.5\%$
• D. $1.5\%$

Answer 1

Answer: (B)

$f^{\prime }=\frac{v}{v-v_s}{f}_0$
$f$ is such that $f=\frac{110}{100}{f}_0$
When $\frac{v}{v-v_s}=\frac{110}{100}$
$100v=110v-110v_s$
$v=11v_s$
When source is received
$\frac{v}{v+v_s}{f}_0=\frac{x}{100}{f}_0$
Putting $v=11v_s$
$\frac{11}{12}\times 100=x$
$x=91.66\%$
$\%$ change $=100-91.66=8.5\%$