A person standing near a railway track hears the whistle of an approaching train as having a frequency of 1600 Hz. The train’s velocity is 40 m/s.

What is the true frequency of the sound emitted by the whistle?

###### Solution.

The Doppler effect is the change in frequency of a wave for an observer. The classical Doppler effect formula is given by

f = f_{0}(c+v_{r})/(c+v_{s}) ,

where f is observed frequency, f_{0} is emitted (true) frequency, c is the velocity of waves in the medium, v_{r} is the velocity of the receiver relative to the medium and v_{s} is the velocity of the source relative to the medium.

We know that the person standing near a railway track so v_{r} = 0.

Also, the velocity of the source is negative because the source is approaching the receiver.

Hence the formula becomes

f = f_{0}*c/(c-v_{s})

or we can rewrite it as

f_{0} = f(1-v_{s}/c)

We take the velocity of sound waves in air as 340 m/s and substitute the values into formula.

f_{0} = 1600*(1-40/340) ≈ 1411.76 Hz