Question

A train moves towards a stationary observer with speed $34m{s}^{-1}$. The train sounds a whistle and its frequency registered by observer is $f_1$. If the speed of the train is reduced to $17m{s}^{-1}$, the frequency registered is $f_2$. If the speed of the sound is $340m{s}^{-1}$, then the ratio $\frac{f_1}{f_2}$ is

• A. $\frac{18}{19}$
• B. $\frac{1}{2}$
• C. $2$
• D. $\frac{19}{18}$

Answer 1

The correct option is D $\frac{19}{18}$

In first case,

$v_s=34m{s}^{-1}$, $v_o=0m{s}^{-1}$, $V=340m{s}^{-1}$

From Doppler's effect in sound waves,

$f_1=\frac{V+0}{V-v_s}f=\frac{340}{340-34}f=\left(10/9\right)\ast f$

in second case,

$v_s=17m{s}^{-1}$, $v_o=0m{s}^{-1}$,

From Doppler's effect in sound waves,

$f_2=\frac{340}{340-17}f=\left(20/19\right)\ast f$

therefore,

$\frac{f_1}{f_2}=\left(10/9\right)/\left(20/19\right)=19/18$

Option "D" is correct.