Question

The apparent frequency of the whistle of an engine changes in the ratio 6:5 as the engine passes a stationary observer. If the velocity of sound is $330m{s}^{-1}$ what is the velocity of the engine?

Answer 1

Let f be the frequency of the sound emitted by the engine and vs be its velocity. The apparent frequency f of the whistle as the engine is approaching the observer is $f^{{}^{\prime }}=f\frac{V}{v-v_s}$

But $v=330m{s}^{-1}$

$\therefore f^{{}^{\prime }}=f\frac{330}{330-v_s}$ ....... (1)

Let f" be the apparent frequency of the sound heard when the engine is moving away from the observer. Then

But $v=330m{s}^{-1}$

$\therefore f^{{}^{\prime \prime }}=f\frac{330}{330+v_s}$ ....... (2)

$\frac{\left(1\right)}{\left(2\right)}\to \frac{f^{{}^{\prime }}}{f^{{}^{\prime \prime }}}=f\frac{330}{330-v_s}+f\left(\frac{330}{330+v_s}\right)$

$\therefore \frac{f^{{}^{\prime }}}{f^{{}^{\prime \prime }}}=\frac{330+v_s}{330-v_s}$

But $\frac{f^{{}^{\prime }}}{f^{{}^{\prime \prime }}}=\frac{6}{5}$ $\therefore \frac{6}{5}=\frac{330+v_s}{330-v_s}$

cross multiplying

$6\times (330-v_s)=5\times (330+v_s)$

$330\times 6-6V_s=5\times 330-5V_s$

$330\times 6-300\times 5=6V_s+5V_s$

$330(6-5)=11V_s$

$330\times 1=11v_s$

$\therefore v_s=\frac{330}{11}=30m{s}^{-1}$