Question

The difference between the apparent frequency of a source of sound as perceived by the observer during its approach and recession is $2\%$ of the frequency of the source. If the speed of sound in air is $ 300\,ms^{-1}$ , the velocity of the source is

• A. $1.5\,ms^{-1}$
• B. $12\,ms^{-1}$
• C. $6\, ms^{-1}$
• D. $3\,ms^{-1}$

Answer 1

The Correct Option is D

When source approaches the observer, the apparent frequency heard by observer is $n'=n\left(\frac{v}{v-v_{s}}\right)$ ...(i) $v_{s}=$ speed of source of sound. During it recession, apparent frequency $n''=n\left(\frac{v}{v +v_{s}}\right)$ ...(ii) Accordingly $n'=n''=\frac{2}{100} n$ (given) $\therefore n\left(\frac{v}{v-v_{s}}\right)=n\left(\frac{v}{v +v_{s}}\right)$ $=\frac{2}{100} n$ or $v\left[\frac{v +v_{s}-v+ v_{s}}{\left(v-v_{s}\right) v +v_{s}}\right]=\frac{2}{100}$ or $\frac{2 v v_{s}}{\left(v-v_{s}\right)\left(v+ v_{s}\right)}=\frac{2}{100}$ or $100\, v v_{s}=v^{2}-v_{s}^{2}$ But speed of sound in air $v=300\, m / s$ $\therefore 30000\, v_{s}=(300)^{2}-v_{s}^{2}$ $\Rightarrow v_{s}^{2}+30000\, v_{s}-90000=0$ $\therefore v_{s}=\frac{-30000 \pm \sqrt{(30000)^{2}+4 \times 90000}}{2}$ $=\frac{-30000 \pm 30006}{2}$ $=\frac{6}{2}=3\, ms ^{-1}$