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Questions about Doppler Effect

  1. Oct 1, 2007 #1
    the formula of Doppler Effect
    f = fs(v + vd)/(v - vs)
    (v = speed of sound; d = detector; s = source)

    Simply from this formula, it can be seen that vd has differenct effect on the frequency receiver from vs.
    when vd or vs approaches the speed of sound, this difference is pronounced.
    make v = 343 m/s, fs = 1 Hz, the the source and detector move towards each other

    Case 1
    vd = 340 m/s, vs = 0
    f = 683/343 = 1.99 Hz
    Case 2
    vd = 0, vs = 340 m/s,
    f = 343/3 = 114 Hz

    why would there be a difference? Isn't the source and detector move relative to each other?
    my teacher says it has sth to do with medium, but i still don't get it.
  2. jcsd
  3. Oct 2, 2007 #2
    Trying to derive the equation you use from base doppler shift ones, not working for me >,< and I only explain stuff now if im 100% sure of where the equations come from, to understand how messy mine is, I have produced:

    [tex]F_{observer} = \frac{V^{2}*F_{source}}{(V-A_{source})(V+B_{observer})} = \frac{V^{2}*F_{Source}}{V^{2}+VB-VA-AB}[/tex]

    A is source velocity
    B is observer velocity
    V is wave speed
    F is Wave frequency

    So yeah, I dont trust my math :)
    Last edited: Oct 2, 2007
  4. Oct 2, 2007 #3
    May aswell wait for a PF guy with more knowledge to pop over the hill annny minute now :)
  5. Oct 2, 2007 #4
    Wrong formula

    Indeed, the formula I wrote here was not correct, therefore I deleted it.
    Last edited: Oct 2, 2007
  6. Oct 2, 2007 #5
    Yeah, what PieWie said, my equation is just a linear combination of the source moving and the observer moving, but that was my attempt to try and show how your irrational equation actually works I guess, since I went off the suggestion that reletive speed wouldnt work, I just combined the equation twice...I really should have just used relative velocity >,<, to rewrite PieWie's equation for easier reading then going explosive on your teacher.

    [tex] f_{observed} = f_{source}\leftbracket[\frac{V}{V+V_{Relative}}\rightbracket] [/tex]

    Thanks PieWie ^_^ I too was wondering why the "compression" (Increase in frequency) was disproportionate to the "decompression" (decrease from moving away from source)
  7. Oct 2, 2007 #6

    Doc Al

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    Staff: Mentor

    To understand why there's a difference, you need to understand how the Doppler formula is derived.

    The fundamental difference is that when you (the detector) remain still but the source moves towards you, the speed of sound with respect to you doesn't change. But the frequency does, since the wavelength is getting shorter. If the original wavelength is [itex]\lambda_0[/itex], the new wavelength is [itex]\lambda = \lambda_0 - v_s / f_0[/itex]. Since for any wave, [itex]v = f\lambda[/itex], the new frequency is:

    [tex]f = v/\lambda = v/(\lambda_0 - v_s / f_0) = f_0 \frac{v}{(v - v_s)}[/tex]

    When the source remains fixed but you (the detector) move towards it, the wavelength stays the same but the speed of sound (relative to you) is now [itex]v' = v + v_d[/itex]. Thus the apparent frequency is:

    [tex]f = v'/\lambda_0 = (v + v_d)/\lambda_0 = f_0 \frac{(v + v_d)}{v}[/tex]

    Of course, if both source and detector are moving, you get the combination of both effects:

    [tex]f = f_0 \frac{(v + v_d)}{(v - v_s)}[/tex]

    I hope that makes a bit of sense.
  8. Oct 2, 2007 #7
    I dont know weather to love doc'al or slap myself in the face :) or do both, so doc'al, how is this handled with EM Radiation? as the speed is constant in that case, would you just use a relative speed case for that?

    EDIT: Oh, and I didnt quite understand:

    [itex]\lambda = \lambda_0 - v_s / f_0[/itex]

    For my low-level ability, it sort of sounds like your treating the sources movement as a wave emmitter in its own right, but taking the wavelength produced from the sound... is it ok for a deeper explanation? You know I'm a picky little mofo.
    Last edited: Oct 2, 2007
  9. Oct 2, 2007 #8
    The [tex]F_{0}[/tex] From the first equation, is equal to the [tex]f[/tex] from the last one, so you substitute the [tex]f_{0}[/tex] in the second equation to give:

    [tex] f_{final}=f_{0}\bracketleft[\frac{v(v+v_{d})}{v(v-v_{s})}\bracketright] [/tex] then cancel variables ^^
  10. Oct 2, 2007 #9

    Doc Al

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    Staff: Mentor

    It turns out that way, but you also have to factor in time dilation. Read this: Relativistic Doppler Effect

    The source is the emitter but I am modifying the wavelength since the source is moving with respect to the medium.
    But cuddly and loveable, I'm sure!
  11. Oct 2, 2007 #10
    And I apologise to doc'al, the equation makes perfect sense, since it was the first one i wrote in my notebook, then I used the hyperphysics one, but let me just clarify:

    [tex] \lambda_{new} = \lambda_{old} - v_{s}*T [/tex]

    and because [tex] T=f^{-1} [/tex] then:

    [tex] \lambda_{new} = \lambda_{old} - \frac{v_{s}}{f_{0}} [/tex]
  12. Oct 2, 2007 #11
    Sorry I didnt get that one, it was glaring me literally in the face in my PF notebook :), also, not to hiijack, but I did some maths in that "Pascals Principle Headache" post invloving gases rather than liquids, I'd really appreciate if you could have a glance over it whenever you have free time
  13. Oct 2, 2007 #12

    Doc Al

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    Staff: Mentor

    Exactly right.

    I'll give it a look a bit later.
  14. Oct 2, 2007 #13
    If the source moves the sound waves have a different frequency in the medium (compared to when the source would stand still or move with a different velocity).

    That is NOT the case for the detector: the frequency in the medium does not depend on the speed of the detector. Note that the frequency that is indicated by the detecor is not equal to the frequency in the medium. So... don't trust a moving sound-detector ;)

    Can someone confirm this explanation?
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