Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Doppler Effect and frequency

  1. Feb 6, 2007 #1
    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. jcsd
  3. Feb 6, 2007 #2
  4. Feb 6, 2007 #3
    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....

    ??
     
  5. Feb 6, 2007 #4
    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Doppler Effect and frequency
  1. Doppler effect? (Replies: 2)

  2. The doppler effect (Replies: 22)

Loading...