Doppler Effect and frequency

1. Feb 6, 2007

studentmom

I have a question regarding the doppler effect and frequency:

You (a stationary observer) are standing by the railroad tracks and hear a frequency of 1000 Hz as the train approaches... as the train goes away, the frequency changes to 800 Hz. Knowing that the speed of sound is 340 m/s, how fast is the train moving?

Now, I understand that the observed frequency seems different (larger) as the train approaches, and smaller when the train leaves. However, I cannot figure out how to find the actual frequency from the source in order to calculate the speed of the train. It seems to me that I have 2 unknowns. The equation I was trying to use was:
f (observed) = f (source) * (speed of sound/
speed of sound -
speed of train)

Any help would be appreciated!

2. Feb 6, 2007

FunkyDwarf

Last edited by a moderator: Apr 22, 2017
3. Feb 6, 2007

studentmom

Thanks, but the information states the same thing that I already knew... the equation is the same, and it offers no explanation of how to figure out the actual frequency of the train's noise....

??

4. Feb 6, 2007

lightarrow

You don't have only one equation, you have two equations: in the first one you put vd as the train's speed, in the second you put -vd.
So you have two equations in the two unknowns vd and f (train's frequency).