The frequency changes by $10\%$ as a sound source approaches a stationary observer with constant speed $v_s$. What would be the percentage change in frequency as the source recedes the observer with the same speed. ... $\%$ Given that $v_s < v$. ($v =$ speed of sound in air)
Diffcult
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as per given situation,
$\frac{\nu^{\prime}}{\nu}=\frac{V}{V-V_{s}}=1.1$
$V_{s}=\frac{1}{11} \times V$
in the second condition,
$\frac{\nu^{\prime}}{\nu}=\frac{V}{V+V_{s}}=\frac{V}{V+(1 / 11) \times V}=\frac{11}{12}$
$\frac{\nu^{\prime}}{\nu} \times 100=\frac{11}{12} \times 100=91.66^{\circ} \%$
hence change is around $8.5 \%$
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