Question

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

• A. 12 ms^-1
• B. 1.5 ms^-1
• C. 3 ms^-1
• D. 6 ms^-1

Answer 1

The correct option is C

3 ms^-1

During approach, source is moving towards listener.

$\therefore v^{\prime }=\frac{v{v}_s}{v-v_s}$

After crossing, source is moving away from listener,

$v^{\prime }=\frac{v{v}_s}{v+v_s}$

$\therefore v^{\prime }-v^{\prime \prime }=v{v}_s(\frac{1}{v-v_s}-\frac{1}{v+v_s})$

$\frac{v^{\prime }-v^{\prime \prime }}{v_s}=v\left(\frac{2v_s}{v^2-v_s^2}\right)$

$\therefore v_s=3m/s[\because v^{\prime }-v^{\prime \prime }=\frac{2v_s}{100}andv=300m/s]$