Question

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

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

Answer 1

Answer: (B)

Apparent frequency heard by the observer for the given case is
$\upsilon'=\upsilon_{0} \left(\frac{\upsilon}{\upsilon-\upsilon_{s}}\right)$
where
$\upsilon_{0}$ = frequency of source $=400\, Hz$
$\upsilon$ =velocity of sound in medium $=1100\,m\, s^{-1}$
$\upsilon_{s}$ =velocity of source $= 100\, m\, s^{-1}$
$\therefore \, \upsilon' =400\left(\frac{1100}{1100-100}\right)$
$=\frac{400\times1100}{1000}$
$=440\,HZ$