Question

A siren placed at a railway platform is emitting a sound of frequency $5kHz$. A passenger sitting in a moving train $A$ records the frequency of the siren as $5.5kHz$. During his return journey by train $B$ he records the frequency of the siren as $6kHz$. The ratio of the speed of train $B$ to that of train $A$ is

• A. $\frac{242}{252}$
• B. $2$
• C. $\frac{5}{6}$
• D. $\frac{11}{6}$

Answer 1

Answer: (B)

Given, frequency of sound, $f_s=5kHz$
frequency of siren records by moving train $A$,
$f_A=5.5kHzE$
frequency of siren records by moving train $B$,
$f_B=6kHz$
Now, let $v_s=$ speed of sound
$\therefore$ For observer $A$, frequency of siren records by train $A$ is given as,
$f_A=f_S\left(\frac{v_S+v_A}{v_S}\right)$
or $5.5=5\left(\frac{v_S+v_A}{v_S}\right)$
or$1.1=1+\frac{v_A}{v_S}$
or $v_A=0.1v_S...(i)$
$\therefore$ For observer $B$, frequency of siren records by train $B$ is given as,
$f_B=f_S\left(\frac{v_S+v_B}{v_S}\right)$
$6=5\left(\frac{v_S+v_B}{v_S}\right)\text{ or }6=5\left(1+\frac{v_B}{v_S}\right)$
$\Rightarrow \frac{6}{5}=1+\frac{v_B}{v_S}$
$\Rightarrow \frac{6}{5}-1=\frac{v_B}{v_S}$
$\Rightarrow \frac{6-5}{5}=\frac{v_B}{v_S}$
$\Rightarrow \frac{1}{5}=\frac{v_B}{v_S}$
$\Rightarrow \frac{v_B}{v_S}=0.2$
or $v_B=0.2v_S...(ii)$
Now, from Eqs. (i) and (ii), we get or
$\frac{v_B}{v_A}=\frac{0.2v_S}{0.1v_S}$ or $\frac{v_B}{v_A}=2$
So, the ratio of the speed of train $B$ to that of train $A$ is $v_B:v_A=2$