Question

An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What will be the percentage increase in the apparent frequency?

• A. Zero
• B. 0.5%
• C. 5%
• D. 20%

Answer 1

The correct option is D 20%

Given :$v_0=\frac{v}{5}\Rightarrow v_0=\frac{320}{5}=64m/s$
When observer moves towards the stationary source, then
$n^{\prime }=\left(\frac{v+v_0}{v}\right)n$
$n^{\prime }=\left(\frac{320+64}{320}\right)n$
$n^{\prime }=\left(\frac{384}{320}\right)n$
$\frac{n^{\prime }}{n}=\frac{384}{320}$
Hence, percentage increase
$\left(\frac{n^{\prime }-n}{n}\right)=(\frac{384-320}{320}\times 100)$%
$=(\frac{64}{320}\times 100)$% $=20$%