Question

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

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

Answer 1

The correct option is D $8.5\%$

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,

$\frac{{\nu }^{\prime }}{\nu }=\frac{V}{V+V_s}=\frac{V}{V+\left(1/11\right)\times V}=\frac{11}{12}$

$\frac{{\nu }^{\prime }}{\nu }\times 100=\frac{11}{12}\times 100=91.66\%$

hence change is around $8.5\%$.

option "D" is correct.