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Doppler Effect

Discussion in 'MCAT Discussions' started by midn, May 12, 2007.

  1. midn

    midn 10+ Year Member

    1,058
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    Sep 30, 2005
    I'm a bit confused with the doppler effect. Apparently, the change in apparent frequency is different if the receiver is moving towards the source or the source is moving towards the receiver at the same speed according to the formula:

    f(D)=f(s)*[(v+v(D))/(v)] (receiver moving towards stationary source)
    f(D)=f(s)*[(v)/(v-v(s))] (receiver stationary; source moving towards receiver)

    where v(s)=v(D)

    I can understand how it is different when using the formula, but this does not make sense to me physically.
     
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  3. Blondnuttyboy

    Blondnuttyboy 7+ Year Member

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    Feb 26, 2007
    If you were to plug in numbers, which one would produce a higher frequency? Moving source or moving receiver?
     
  4. TheFreakMD

    TheFreakMD New Member

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    Aug 19, 2006
    The key thing (and the more logical point for that matter) is to think about all those problems in terms of frequency change. Obviously if you are moving away or losing ground to the sound source, the frequency of the sound is going to decrease. From there, you know that frequency and wavelength are inversely related so-----as frequency increases, wavelength decreases.

    Just practice one problem with the source and the subject moving in all possible directions with respect to each other and you'll see that it's all logical. But of course, MCAT wants you to come up with a specific number so you have to learn the formulas.

    Hope that helps:rolleyes:
     
  5. Maxprime

    Maxprime Higgs chaser 5+ Year Member

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    Nov 28, 2005
    First off, always use real-world examples when you get confused about what happens. When an ambulance passes, think about how the pitch goes up and when it passes you it goes down quickly.

    Another thing to think of is from optics. When a wave changes medium and its velocity changes, wavelength stays the same but frequency changes. Here, your relative velocity of the wave is decreasing as the ambulance drives away - so frequency drops with it and it sounds that way.
     
  6. Foghorn

    Foghorn Guest

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    Feb 13, 2007
    f' = apparent frequency measure by the observer.

    I just keep in mind that velocity is a vector and/or which direction(s) the observer and/or source are moving rellative to each other.

    [​IMG]
     
  7. scottj72

    scottj72 2+ Year Member

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    Apr 3, 2007
    I just wanted to make a few comments on what maxprime said.

    "Here, your relative velocity of the wave is decreasing as the ambulance drives away - so frequency drops with it and it sounds that way."

    Actually the velocity of the wave is unchanged. The wave is always traveling at the speed of sound. Like ThefreakMD stated the frequency is what changes. The change in frequency relates to a change in wavelength because the velocity is constant. The pitch change is due to the change in frequency.

    "Another thing to think of is from optics. When a wave changes medium and its velocity changes, wavelength stays the same but frequency changes."

    This is referring to refraction. This is when the velocity of a wave changes as it moves from one medium to another. Actually it is the frequency stays the same and the wavelength that changes.
     
  8. Vanguard23

    Vanguard23 5+ Year Member

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    Sep 5, 2006
    I think you have that backwards. I think frequency stays the same and wavelength changes.
     
  9. BrokenGlass

    BrokenGlass 2+ Year Member

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    Jan 12, 2007

    http://forums.studentdoctor.net/showpost.php?p=4915900&postcount=806
     
  10. oxeye

    oxeye Moderator Emeritus 7+ Year Member

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    Feb 12, 2006
    Try thinking about it as waves in the middle of the ocean. If you're on boat travelling into the waves, you're going to hit the crests more often which is an apparent increase in frequency relative to the frequency of the waves if you were stationary.

    If you travel away from the waves, you're going to hit the crests less often - so a decrease in frequency.

    It works the same for any wave (sound, light, etc).

    Anyway, I'm not sure if that helps you but when I was learning this concept, it helped to apply it to a situation that I can see in my head rather than think about it in terms of sound waves.
     
  11. Creightonite

    Creightonite Senior Member 7+ Year Member

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    Jun 26, 2003
    is it really required to know Dopplet formulas for MCAT? I know the frequency differences, but cant keep the formulas straight.
     
  12. midn

    midn 10+ Year Member

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    Sep 30, 2005
    Err, thanks everyone for reiterating the Doppler effect numerous times, but you all missed my original question. I know how the Doppler effect works and why it does, but WHY does it make a difference if a receiver is stationary and the source moves towards it or if the source is stationary and the receiver moves towards the source at the same speed?

    I mean, couldn't you technically mess with frames of reference to say that, although you are moving towards the source, in one frame of reference, the source is moving towards you?

    Going back to my original formulas:

    f(D)=f(s)*[(v+v(D))/(v)] (receiver moving towards stationary source)
    f(D)=f(s)*[(v)/(v-v(s))] (receiver stationary; source moving towards receiver)

    where v(s)=v(D)

    plug in v(s)=v(D)=330m/s and say speed of sound is v=340m/s

    f(D) in the first case would be about 2*f(s) while f(D) in the second case is 34*f(s). What gives?
     
  13. Foghorn

    Foghorn Guest

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    Feb 13, 2007
    Because when the source/observer is moving away/towards the observer/source, the distance between the wavefronts of the emitted sound changes. As a consequence the frequency, as perceived by the observer is either higher/lower.
     
  14. mrmilad

    mrmilad 2+ Year Member

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    Jun 7, 2007
    MDApps:
    Hello everyone, I wanted to ask if using the ExamCrackers watered down equations of:

    delta f / fs = v / c

    and

    delta waveLngth / waveLngth s = v / c

    Would be acceptable for the mcat? I understand how the positioning and the velocity changes occur, and it appears to me that these equations are compatible to almost all the doppler questions I have seen. Are these fine?
    Albeit that these equations require a qualitative understanding to find the direction of frequency change.
     
  15. DiscoDoc

    DiscoDoc Till I Collapse 2+ Year Member

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    Jun 26, 2006
    Grenada, West Indies
    I think so. It gives a decent answer with every problem I've tried.
     
  16. axp107

    axp107 UCLA 09': Italian Pryde 2+ Year Member

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    Dec 25, 2006
    Phoenix, AZ
    I don't understand how c, the speed of light just appears in there lol
     
  17. minah86

    minah86 Senior Member 10+ Year Member

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    Aug 21, 2005
    ^ It doesn't have to be the speed of light. It could be the speed of sound in air.
     
  18. corbis11

    corbis11 10+ Year Member

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    Mar 27, 2005
    Philadelphia
    deleted
     

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