Question

A siren placed at a railway platform is emitting a sound of frequency $5\, kHz$. A passenger sitting in a moving train $A$ records the frequency of the siren as $5.5\, kHz$. During his return journey by train $B$ he records the frequency of the siren as $6\, kHz$. 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, $f_{s}=5 \,kHz$
frequency of siren records by moving train $A$,
$f_{A}=5.5 \,kHzE$
frequency of siren records by moving train $B$,
$f_{B}=6 \,kHz$
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 $v_{A}=0.1\, v_{S}\,\,\,...(i)$
$\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 $v_{B}=0.2\, v_{S}\,\,\,...(ii)$
Now, from Eqs. (i) and (ii), we get or
$\frac{v_{B}}{v_{A}}=\frac{0.2\, v_{S}}{0.1 \,v_{S}}$ or $\frac{v_{B}}{v_{A}}=2$
So, the ratio of the speed of train $B$ to that of train $A$ is $v_{B}: v_{A}=2$