A siren placed at a railway platform is emitting a sound of frequency $5 \mathrm{kHz}$. A passenger sitting…
A siren placed at a railway platform is emitting a sound of frequency $5 \mathrm{kHz}$. A passenger sitting in a moving train $A$ records the frequency of the siren as $5.5 \mathrm{kHz}$.
During his return journey by $\operatorname{train} B$ he records the frequency of the siren as $6 \mathrm{kHz}$. The ratio of the speed of $\operatorname{train} B$ to that of $\operatorname{train} A$ is
$\frac{242}{252}$
2
$\frac{5}{6}$
$\frac{11}{6}$
Solution
Given, frequency of sound, $f_s=5 \mathrm{kHz}$ frequency of siren records by moving train $A$,
$
f_A=5.5 \mathrm{kHz}
$
frequency of siren records by moving train $B$,
$
f_B=6 \mathrm{kHz}
$
Now, let $v_s=$ speed of sound
$\therefore$ For observer $A$, frequency of siren records by train $A$ is given as,
$
\begin{aligned}
& f_A & =f_S\left(\frac{v_S+v_A}{v_S}\right) \\
\text { or } & 5.5 & =5\left(\frac{v_S+v_A}{v_S}\right) \\
\text { or } & 1.1 & =1+\frac{v_A}{v_S}
\end{aligned}
$
$\therefore$ For observer $B$, frequency of siren records by train $B$ is given as,
$
\begin{array}{cc}
& f_B=f_S\left(\frac{v_S+v_B}{v_S}\right) \\
\Rightarrow \quad 6=5\left(\frac{v_S+v_B}{v_S}\right) \text { or } 6=5\left(1+\frac{v_B}{v_S}\right) \\
\Rightarrow \quad \frac{6}{5}=1+\frac{v_B}{v_S} \\
\Rightarrow \quad \frac{6}{5}-1=\frac{v_B}{v_S} \Rightarrow \frac{6-5}{5}=\frac{v_B}{v_S} \\
\frac{1}{5}=\frac{v_B}{v_S} \Rightarrow \frac{v_B}{v_S}=0.2
\end{array}
$
Now, from Eqs. (i) and (ii), we get or
$
\frac{v_B}{v_A}=\frac{0.2 v_S}{0.1 v_S} \text { or } \frac{v_B}{v_A}=2
$
So, the ratio of the speed of $\operatorname{train} B$ to that of train $A$ is $v_B: v_A=2$
