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B Doppler effect formula confusion

  1. Apr 24, 2017 #1
    Hello! I am a bit confused by the Doppler effect. Based on the formula it seems that if the source moves towards me with speed ##v## I hear a frequency ##\nu_1## but if I move towards the source with the same speed ##v## I hear a different frequency ##\nu_2##. However according to the equivalence of inertial reference frame (both classically and relativistically) I should not be able to tell if I am moving towards the source or if the source moves towards me (and based on doppler effect I can do an experiment to tell me if I am stationary or moving with constant velocity). So what am I missing here? Thank you!
     
  2. jcsd
  3. Apr 24, 2017 #2

    russ_watters

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    Can you post the equation you are using? You are probably misunderstanding it...there should only be one speed in it besides the speed of sound.
     
  4. Apr 24, 2017 #3
    Assume that the speed of sound is ##c## and the frequency of the source is ##\nu_0##. If I move towards the source with the speed ##v## and the source stand still, the frequency I hear is ##\nu_1 = \frac{c+v}{c}\nu_0## if the source moves towards me with the same speed ##v## and I stay still the frequency I hear is ##\nu_2 = \frac{c}{c-v}\nu_0##. ##\nu_1## and ##\nu_2## are not the same, so I can tell if it is me or the source that is moving.
     
  5. Apr 24, 2017 #4
    The two situations are not the same. The difference is accounted for by the part the air, the media of propagation, plays. In the moving source scenario the source motion modifies the wavelength of the sound the observer hears the speed of propagation not being affected. In the case of the moving observer the wavelength is unaffected but the frequency is modified because the observer hears different cycles per second with is proportional to the relative velocity. of the observer and the sound.
     
  6. Apr 24, 2017 #5
    Wait, I am not sure I understand. In both situations, the speed of propagation is unchanged (it is equal to c). And we have that ##\lambda = c \nu##. So in both cases the frequency (and hence the wavelength) is changed, just by a different amount. So what do you mean by "the wavelength is unaffected but the frequency is modified"? You can't have a change in frequency, without a change in wavelength.
     
  7. Apr 24, 2017 #6
    Sorry I missed the object of your OP.I was describing the acoustical DE.
     
  8. Apr 24, 2017 #7

    A.T.

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    ... relative to the medium (air), not in an absolute sense. You don't need Doppler to determine that.
     
  9. Apr 24, 2017 #8
    I am not sure I understand. My question is, shouldn't the physics be the same in 2 inertial reference frames? For example in relativity, if 2 observers have a meter stick and one of them is moving and the other is stationary, the length of the meter stick saw from the other frame is the same for both of them (i.e. observer 1 sees that the meter stick of observer 2 has length L' but also observer 2 sees the length of observer 1 to be L') so they can't tell which one is moving (relative to a fixed reference frame). But in the case of Doppler effect, assuming that for example the street is the stationary reference frame I can tell if the car is moving towards me or I am moving towards the car (relative to the street).
     
  10. Apr 24, 2017 #9

    A.T.

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    The physics is the same. The velocity of the air mass isn't, so neither is the sound propagation.

    That's a symmetrical situation, with two meter sticks, not with one air mass.
     
  11. Apr 24, 2017 #10
    What do you mean by air mass? What does that have to do with frequency of the sound?
     
  12. Apr 24, 2017 #11

    A.T.

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  13. Apr 24, 2017 #12
    OP, your question is a good one, and one that has been asked here before as well as elsewhere. In trying to find a good explanation for you, I came across a post on another forum, credited below, that I will elaborate on here.

    Suppose you have three bodies, from left to right A, B and C in a frame of reference in which they are all at rest and consider two cases. In the first case, A starts moving at velocity v to the right, so it is approaching B. In the second case, instead have B move with velocity v to the left, approaching A. In both cases, A and B are approaching each other with the same relative velocity. So you would say these two cases are the same and all effects must be symmetric.

    But what about C? In the first case, A is approaching C with velocity v and in the second case A is stationary with respect to C. So the two cases are not the same and if there is an effect that depends on the relative velocity of A and C, the results would be different. To relate this to the Doppler effect, let A = source, B = observer and C = air.

    https://physics.stackexchange.com/a/257041
     
  14. Apr 25, 2017 #13

    A.T.

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