Question

A source of sound $S$ is moving with a velocity of $50 \, m \, s^{- 1}$ towards a stationary observer. The observer measures the frequency of the source as $1000 \, Hz$ . 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 $350 \, m \, s^{- 1}$ .

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

Answer 1

Answer: (A)

Here, $u_{s}=50ms^{- 1}$ , $v_{L}=0, \, u=350ms^{- 1}$
When the source is moving towards the observer,
$\nu^{′}=1000$
$\nu^{′}=\frac{u \times \nu}{u - u_{s}}$
$\nu=\frac{\left(u - u_{s}\right) \left(\nu\right)^{′}}{u}$
$=\frac{\left(\right. 350 - 50 \left.\right) 1000}{350}=\frac{6000}{7}Hz$
When the source is moving away from the observer,
$\nu^{′}=\frac{u \times \nu}{u + v_{s}}$
$=\frac{350}{\left(\right. 350 + 50 \left.\right)}\times \frac{6000}{7}$
$=750 \, Hz$