Question

When the observer moves towards the stationary source with velocity, $V_1$, the apparent frequency of emitted note is $F_1$. When the observer moves away from the source with velocity $V_1$, the apparent frequency is $F_2$. If $V$ is the velocity of sound in air and $\frac{F_{1}}{F_{2}} = 2$ then $\frac{V}{V_{1}}=$ ?

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The Correct Option is B

$F_{1}=\left(\frac{v+v_{1}}{v}\right) f$
where $f$ is the real frequency
and $F_{2}=\left(\frac{v-v_{f}}{v}\right) f$
$\therefore \frac{F_{1}}{F_{2}}=\frac{2}{1}=\frac{v+v_{1}}{v-v_{1}}$
$\therefore 2 v-2 v_{1}=v+v_{1} $
$\therefore v=3 v_{1}$
$\therefore \frac{v}{v_{1}}=3$