Question

An observer moves towards a stationary source of sound with a velocity $\frac{1}{20}$th of the velocity of sound in air. The apparent increase in frequency is?

• A. $5\%$
• B. $10\%$
• C. $3\%$
• D. $8\%$

Answer 1

The correct option is A $5\%$

When an observer is moving towards a stationary source of sound then the apparent frequency is given by
$f^{\prime }=f\left(\frac{v+v_0}{v}\right)=f\left(1+\frac{v_0}{v}\right)$
Here, velocity of observer $=v_0=\frac{1}{20}v$
So, $f^{\prime }=f\left(1+\frac{v_0}{20v_0}\right)=f\left(1+\frac{1}{20}\right)$
$\therefore f^{\prime }-f=\frac{1}{20}f$
Thus, the apparent increase in frequency is
$\frac{f^{\prime }-f}{f}\times 100=\frac{\frac{1}{20}f}{f}\times 100=5\%$.