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

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,
\(\frac{v}{v_{2}}=\frac{5}{4}=1.25\)