Question: Two trains emit \({\bf{508 - Hz}}\) whistles. One train is stationary. The conductor on the stationary train hears a \({\bf{3}}{\bf{.5 - Hz}}\) beat frequency when the other train approaches. What is the speed of the moving train?

Doppler Effect: The apparent frequency changes when the listener and the source both have a relative motion.The apparent frequency is given as,

\({f^'} = f\frac{{{v_{\rm{s}}}}}{{\left( {{v_{\rm{s}}} - {v_{{\rm{source}}}}} \right)}}\)

Here,\(f\) is the actual frequency emitted from the source,\({v_s}\) is the speed of sound in air,\({v_{source}}\) is the speed of the source.

The frequency of the whistle is\(f = 508\;{\rm{Hz}}\).

The beat frequency is\(\Delta f = 3.5\;{\rm{Hz}}\).

The train is approaching towards the stationary train, so the apparent frequency seems to be greater than the actual frequency.

The apparent frequency is,

\(\begin{array}{c}f' = f + \Delta f\\ = 508\;{\rm{Hz}} + 3.5\;{\rm{Hz}}\\ = 511.5\;{\rm{Hz}}\end{array}\)

The speed of sound in air is\({v_s} = 343\;{\rm{m/s}}\).

The apparent frequency is,

This implies,

Substitute the values and get,

\(\begin{array}{c}{v_{{\rm{source}}}} = \left( {1 - \frac{{508\;{\rm{Hz}}}}{{511.5\;{\rm{Hz}}}}} \right) \times 343\;{\rm{m/s}}\\{v_{{\rm{source}}}} \approx 2.35\;{\rm{m/s}}\end{array}\)

Hence, the speed of the train is\(2.35\;{\rm{m/s}}\).