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The doppler effect and EM waves

  1. Mar 6, 2007 #1
    1. The problem statement, all variables and given/known dataSuppose that a police car on the highway is moving to the right at 28 m/s, while a speeder is coming up from almost directly behind at a speed of 34 m/s, both speeds being with respect to the ground. The police officer aims a radar gun at the speeder. Assume that the electromagnetic wave emitted by the radar gun has a frequency of 10.0x10^9 Hz.

    * Find the magnitude of the difference between the frequency of the emitted wave and the wave that returns to the police car after reflecting from the speeder's car.

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    2. Relevant equations
    doppler Effect equation



    3. The attempt at a solution i know the doppler equation, but where i am confused is relative velocity. i have tried working this problem out numerous ways: i have subtracted to the two speeds, i have added their speeds, i dont know how to come to the right conclusion in terms of relative speed. i noticed my book says that "do not use the speed of the observer or of the source with respect to the ground in the doppler equation." If that is the case, then what am i supposed to use??? Also, one object is moving away, while one is moving towards the other. Do i use the plus or minus sign in the doppler equation. i would say the + sign since the speeder is moving faster than the cop car and will eventually pass it.
     
    Last edited: Mar 6, 2007
  2. jcsd
  3. Mar 6, 2007 #2

    Dick

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    You should use the difference of the two speeds. And, yes, since they are closing on each other, the sign should be +.
     
  4. Mar 6, 2007 #3
    Dick, i did what you suggested but the answer i get is not right.
    using the doppler equation to solve for the observed frequency:
    fs*[1+(6m/s / speed of light) = 1.0x10^10 Hz
    Now since they ask for the difference between the frequency of the emitted wave and the wave that returns to the police car after reflecting from the speeder's car i would do the following:
    10.0x10^9 Hz - 1.0x10^10 = 0 !!!
    Can you see anything off hand that i am screwing up here?
     
  5. Mar 6, 2007 #4

    Mentz114

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    You need to calculate to about 9 decimal places before the difference is apparent. Don't use 3x10^8 M/s because it's not accurate enough.
    it's actually 2.994.... you'll need to look it up.

    Oh, you may also have to double the velocity difference because the wave is being dopplered twice (?)
     
    Last edited: Mar 6, 2007
  6. Mar 7, 2007 #5

    Dick

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    Do use more decimal places. You don't really need more accuracy in c, it won't make that much difference in the ratio. And yes, I think there is a factor of two since it's a reflected signal.
     
    Last edited: Mar 7, 2007
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