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Doppler effect from an angle

  1. May 16, 2009 #1
    Doppler Effect- Observer On Angle To Source

    Plz help- I've been working on this for 2 days and I'm down to my last teaspoon of coffee powder. Even having the answer didn't help!
    I think sleepness is turning me into a zombie :eek: ...prehaps eating brains will help me solve this problem...

    An ambulance is traveling at 84.0 km/h towards an intersection. Jim stops his car to give way to the ambulance as shown in the diagram. θ = 42.0 °.
    http://oasis.phy.auckland.ac.nz/oasis/a/att/qtatt/888/2/52/amburightjimrest.gif
    The ambulance has a siren which produces sound at a frequency of 2.8 kHz. Assume the speed of sound in air is 340.0 m/s. What is the doppler shift (Δf = f′–f) heard in the siren sound by Jim, when Jim and the ambulance are positioned as shown on the diagram above?
    Ans: 0.148 kHz

    3. The attempt at a solution
    I figured f' =f [ v/(v-v(source) ] then Δf = f′–f but no luck there :frown:
    Does it have something to do with the angle?
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. May 16, 2009 #2

    Doc Al

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    Staff: Mentor

    Re: Doppler Effect- Observer On Angle To Source

    There's your problem--you need some real coffee.
    Sure. What's the component of the ambulance's speed toward Jim?
     
  4. May 16, 2009 #3
    Either this is a very interesting question or I am doing something wrong.

    I got 0.147 making an assumption. I had to make it because I think the doppler shift in this case depends on the distance between the source and the receiver. If you draw the geometry, you will see that there is some line segment that you need to figure out the length of to get the answer. And to do that you either make an assumption (that should not make a difference if that not-given distance is indeed not needed), or you need that distance I mentioned to be given.

    Just consider where the ambulance will be when the sound signal has travelled one wavelength.

    Anyways, just tell me what you think. I might be wrong about the assumption.
     
  5. May 16, 2009 #4
    Thanks guys, although I'm still a little confused. I can work out the values, but I'm not sure what to do with them- it's like trying to piece together a jigsaw puzzle without the picture on the box lid. Any more clues?

    Your answer is probably right- all the values were probably rounded to 3 s.f throughout the calculation.
     
  6. May 17, 2009 #5
    Ok, now I'm trying f'= fsourcecosA/v
    so: f' = 2.8 (23.3cos42)/340= 0.143
    ... which is wrong, but it's the closest I've got so far. Am I at least on the right track?
     
    Last edited: May 17, 2009
  7. May 17, 2009 #6

    Doc Al

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    I don't understand what you're doing. Use your original Doppler formula, only with the component of the ambulance's speed toward Jim instead of its full speed.
     
  8. May 21, 2009 #7
    the component of the ambulance's speed toward Jim instead of its full speed, and use that in the doppler shift equation, I think you get 0,145.
    however, if you
    1) draw the geometry (i.e. draw where the ambulance will be after the wave has travelled one wavelength)
    2) draw the line segment that connects the ambulance and Jim at this point in time (after step 1)
    3) and try to calculate the length of the piece between the ambulance and the first wave that has already travelled one wavelength (this distance is the length of only some portion of that line segment drawn in step 2)

    For this calculation, you either need the distance between the source and the receiver, or you assume that Jim is located at the point where the first wave intersects the line segment between the original position of the ambulance and Jim. If that distance is not needed, this assumption should not change the result. Because in this case, it would mean that Jim's location is not important as long as he is somewhere on that line segment.

    Then you use law of cosines to figure out the distance and do a quick ratio to get a corresponding frequency. That's how I got 0,147.

    I know this is extremely confusing to read and understand. I'll get a picture as soon as I find a scanner.
     
  9. May 21, 2009 #8

    Doc Al

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    I get closer to 0.150 kHz. (But I haven't had my coffee yet.)
     
  10. May 21, 2009 #9
    oh I'm sorry. Yes, if I round it to 3 decimal places, I think it is 0,150kHz.
    But is that right?

    I haven't had my coffee either :)
     
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