A train blowing the whistle moves with a constant velocity ' $\mathrm{V}$ ' away from an observer standing…
A train blowing the whistle moves with a constant velocity ' $\mathrm{V}$ ' away from an observer standing on the platform. The ratio of the natural frequency of the
whistle "n' to the apparent frequency is $1 \cdot 2: 1$. If the train is at rest and the
observer moves away from it at the same velocity 'V', the ratio of 'n' to the
apparent frequency is
$0.51: 1$
$1 \cdot 25: 1$
$2 \cdot 05: 1$
$1 \cdot 52: 1$
Solution
If the train is going away from the observer, the apparent frequency is
$\mathrm{v}_{1}=\frac{\mathrm{vu}}{\mathrm{v}+\mathrm{u}}=\frac{\mathrm{v}}{1+\frac{\mathrm{u}}{\mathrm{v}}}$
It is observed that $\mathrm{v}=1.2 \mathrm{v}_{1}$ (Given),
In the second case the apparent frequency is
$\mathrm{v}_{2}=\frac{\mathrm{v}(\mathrm{v}-\mathrm{u})}{\mathrm{v}}=\mathrm{v}\left(1-\frac{\mathrm{u}}{\mathrm{v}}\right)$
or
$\frac{\mathrm{v}}{\mathrm{v}_{2}}=\frac{1}{1-\frac{\mathrm{M}}{\mathrm{v}}}$
Now, from equation (1) we have
$\frac{v}{v_{1}}=1+\frac{u}{v}$
or
$1.2=1+\frac{\mathrm{u}}{\mathrm{v}}$
$\mathrm{u}=0.2 \mathrm{v}$
That is,
$\frac{u}{v}=0.2$
Using this in equation (2), we get,
$\frac{v}{v_{2}}=\frac{5}{4}=1.25$