Doppler Effect

Trains travel at speeds less than the speed of sound. Because of this, there will be sound waves, from the train whistle, in front of the train that are compressed somewhat because the train is moving. They will have a pitch that is higher than the pitch you hear if the train is not moving.

Also, sound waves moving behind the train are expanded somewhat because the train is moving. They will have a pitch that is lower than the pitch you hear if the train is not moving.

In the equation below, the math is set up so that a positive value for the source (the train) corresponds to the train moving towards you and a a positive value for you means you are moving towards the train.

  • v_{wave} = speed of sound, 343 m/s
  • v_{source} = speed of the train (m/s)
  • v_{observer} = speed of you (m/s) // in many problems this will be 0 m/s
  • f_{actual} = the frequency heard when the train isn't moving
  • f_{apparent} = the frequency you hear

f_{apparent} = f_{actual} \dfrac {v_{wave} + v_{observer}} {v_{wave} - v_{source}}

Example problem: the speed of sound is 340 m/s, the train is moving 35 m/s and the frequency of the train whistle is 150 MHz. Calculate the frequency of the sound you hear when you briefly stand on the tracks in front of the train, and then calculate the frequency of the sound you hear when you stand on the tracks again while the train is moving past you.

We do the calculation, taking care to put in “0” for v_observer. We calculate that when the train is moving towards us we will hear a frequency of 167 MHz and when the train is moving away from us we will hear a frequency of 136 MHz.

Now, let’s change the problem a bit. We will still have the direction of positive x facing to the right and we will have the train in a location to the right of the observer (you).

But we change it so if the train is moving to the right (away from you), the velocity will be positive and if the train is moving to the left (toward you) the velocity will be negative.

f_{apparent} = f_{actual} \dfrac {v_{wave} + v_{observer}} {v_{wave} + v_{source}}

You might say the change we made to the problem and the change we made to the calculation, canceled each other. You should still get 167 MHz for “moving towards me” and 136 MHz for “moving away from me”.

We are grateful to the author of the web page for giving us discussion and a simple problem with numbers (thank you!) but when we read the part about having two separate equations for “towards” and “away from”, it sounded like fingernails on the chalkboard.