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Apparent Frequency (Doppler) Problem

  1. Feb 25, 2009 #1
    The security alarm on a parked car goes off and produces a frequency of 952 Hz. The speed of sound is 343 m/s. As you drive toward this parked car, pass it, and drive away, you observe the frequency to change by 97 Hz. At what speed are you driving?

    F=Fo [(Vsound- Vobserver)/(Vsound-Vsource)]


    855= 952 [(343-V)/(343-0)]

    V= 34.9485 m/s
  2. jcsd
  3. Feb 25, 2009 #2
    I think you're maybe missing a part of the problem...

    On your approach, is the frequency is higher, or lower? What about when you're going away from the source?

    So, if you notice a difference of 97 Hz, what is it the difference of - the frequencies heard at which points in your travel? What are your velocities during these times, in relation to the source?
  4. Feb 25, 2009 #3
    The question only asks for one speed, and I added the 97 and subtracted 97 and it was still wrong. And its when you are moving away from the parked car, so wouldn't it be subtracted, anyway? And you're solving for the velocity when you're moving away.

    So all in all, I have no idea what you just said
  5. Feb 25, 2009 #4
    Okay, let me try again...

    What's the frequency when you're aproaching the siren? What's the frequency when you're going away from it? The difference in frequency should be between these two numbers, I think...

    Basically, try for something like [ F(approaching) - F(going away) ] = 97 Hz
  6. Mar 5, 2009 #5
    Okay, since 97 is the difference for approaching AND moving away from the car, you only want half the distance. That should solve you problem :)

    (Just divide the answer you got by 2)
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