Question

A source of sound $S$ is moving with a velocity of $50m{s}^{-1}$ towards a stationary observer. The observer measures the frequency of the source as $1000Hz$ . What will be the apparent frequency of the source when it is moving away from the observer after crossing him? The velocity of the sound in the medium is $350m{s}^{-1}$ .

• A. $750Hz$
• B. $857Hz$
• C. $1143Hz$
• D. $1333Hz$

Answer 1

Answer: (A)

Here, $u_s=50m{s}^{-1}$ , $v_L=0,u=350m{s}^{-1}$
When the source is moving towards the observer,
${\nu }^{\prime }=1000$
${\nu }^{\prime }=\frac{u\times \nu }{u-u_s}$
$\nu =\frac{\left(u-u_s\right){\left(\nu \right)}^{\prime }}{u}$
$=\frac{(350-50)1000}{350}=\frac{6000}{7}Hz$
When the source is moving away from the observer,
${\nu }^{\prime }=\frac{u\times \nu }{u+v_s}$
$=\frac{350}{(350+50)}\times \frac{6000}{7}$
$=750Hz$