Question

A source of sound of frequency $400Hz$ is moving with a velocity of $100m{s}^{-1}$ towards a stationary observer. If the velocity of sound in the medium is $1100m{s}^{-1}$, then the frequency heard by the observer will be

• A. $400Hz$
• B. $440Hz$
• C. $480Hz$
• D. $460Hz$

Answer 1

Answer: (B)

Apparent frequency heard by the observer for the given case is
${\upsilon }^{\prime }={\upsilon }_0\left(\frac{\upsilon }{\upsilon -{\upsilon }_s}\right)$
where
${\upsilon }_0$ = frequency of source $=400Hz$
$\upsilon$ =velocity of sound in medium $=1100m{s}^{-1}$
${\upsilon }_s$ =velocity of source $=100m{s}^{-1}$
$\therefore {\upsilon }^{\prime }=400\left(\frac{1100}{1100-100}\right)$
$=\frac{400\times 1100}{1000}$
$=440HZ$