Question

A train is passing by a platform at a constant speed of $40m/s$. The train horn is sounded at a frequency of $378Hz$. Find the overall change in frequency (in $Hz$) as detected by a person standing on the platform, as the train moves from approaching to receding. (Take velocity of sound in air as $320m/s$)

Answer 1

Given,
speed of the train, $v_s=40m/s$
frequency of the train, $n=378Hz$
We know that, speed of the sound, $320m/s$

Speed of the observer (person standing in platform), $v_0=0$

For situation $1$, train is approaching the person,

$\Rightarrow n_1=n\left(\frac{v-v_0}{v-v_s}\right)$

$\Rightarrow n_1=378\times \left(\frac{320-0}{320-40}\right)$

$\Rightarrow n_1=378\times 1.142$

$\Rightarrow n_1=432Hz$

For situation $2$. Train is receeding the person,

$\Rightarrow n_2=n\left(\frac{v-v_0}{v+v_s}\right)$

$\Rightarrow n_2=378\times \left(\frac{320-0}{320+40}\right)$

$\Rightarrow n_2=378\times 0.888$

$\Rightarrow n_2=336Hz$

$\Rightarrow n=n_1-n_2$

$\Rightarrow n=432-336$

$\Rightarrow n=96Hz$

Final answer: $96Hz$