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Frequency of moving source

  1. Feb 6, 2015 #1
    1. The problem statement, all variables and given/known data
    A sound source has frequency 1000Hz and moves at velocity v along a straight line. A person stands a perpendicular distance d away from the line.
    At time t=0 the source is at its distance of closest approach to the listener. Express the frequency as heard by the listener as a function of time.
    2. Relevant equations
    v=fλ

    3. The attempt at a solution
    I can't work out how the wavelength is affected! When the source moves directly towards the listener then the listener hears λl = λs±Tsvs. So I have to do something involving angles, but I really don't know what! I think because of the definition of t=0, the ± won't be necessary in my expression, it will take care of itself.
    Is the velocity of the wave approaching the listener vscosθ, if θ is the angle between the velocity of the source and a line drawn from source to listener?
     
  2. jcsd
  3. Feb 6, 2015 #2

    BvU

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    No ! The velocity of the wave is the speed of sound ! What you describe is the speed of the source in the direction of the observer. It's a good thing to calculate, though, because you need it in a relevant equation (for the frequency) that you haven't listed yet.
    The cosine thing is the right thing to do. And it will indeed deal with the sign in the proper way !
    You're as good as there
     
  4. Feb 6, 2015 #3
    The wavelength for the listener then is the wavelength of the source λs+Tsvs where T is the period of the source.
    λLs+Ts(vscosθ), rearrange a bit with T=1/f and λ=v/f to get

    λL= fsv/(v+vscosθ)
    But now how do I get that in terms of time??
     
  5. Feb 6, 2015 #4

    BvU

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    Cheack your rearrangements: on the right you have a frequency (which is what is asked for). So on the left you can't have a ##\lambda##.

    How to get it in time terms ?
    Easy ! you write ##\theta## as a function of time !
     
  6. Feb 6, 2015 #5
    Is cosθ = d/vt then? And my expression should be

    v/fL=fsv/(v+vscosθ)

    So fL = (v+vscosθ) / fs ?
     
  7. Feb 6, 2015 #6
    If that's right, then I had the wrong theta in mind!
     
  8. Feb 6, 2015 #7

    BvU

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    You defined ##\theta## in post #1. d/(vt) is not the cosine, but the ..., so you are getting closer.

    And it hurts me to see things like
    please please recheck you re-rearrangements.

    And: why not work with ##f = {c-v_s\over c} \; f_0 ## like they do here ?

    Your equation ##\lambda_l = \lambda_s - Tv_s## is easily rewritten using your ##v = \lambda f## to $$
    {c\over f} = {c\over f_0} - {v_s\over f_0}\ \Rightarrow\quad {f_0\over f} = {c - v_s\over c} \ \Rightarrow\quad f = {c-v_s\over c} \; f_0
    $$

    [edit, days later (!)] which of course should be
    $$
    {c\over f} = {c\over f_0} - {v_s\over f_0}\ \Rightarrow\quad {f_0\over f} = {c - v_s\over c} \ \Rightarrow\quad f = {c\over c-v_s} \; f_0
    $$I went a little too fast and looked too much at the first-order approximation in Wiki.
    And vs is positive if the source is moving towards the listener. (See post #12).
    My apologies
    :oops: Someone should have put me right !



    where I already put in the right sign for ##v_s## (but you have to check !)

    Finishing touch: the link reminds me that ##f## is not equal to ##f_0## at t=0, but a little later. How much later ?
     
    Last edited: Feb 8, 2015
  9. Feb 6, 2015 #8
    I do have a frequency on the right there, but it isn't the one the question asks for...
    As to checking my re-arrangement, I'm not surprised that hurts. What a weird way to rearrange. I think
    λL=(v+vscosθ)/fs

    So
    v/fL=(v+vcosθ)/fs

    fL= vfs/(v+vscosθ)

    But I'm not particularly sure I trust my algebra anymore. My expression for cosθ is actually an expression for sinθ, but that's because I thought t cancelled in the expression for cosθ, which I think is vst/vt?

    Thank you for being patient and helping, I'm not sure why this is so confusing to me!
     
    Last edited: Feb 6, 2015
  10. Feb 6, 2015 #9
    I suppose cosθ could be expressed like this instead
    vst/√(vs2t2+d2

    As for the negative, my sign has been wrong from the start then! Going back to
    λl = λs±Tsvs, I want a negative sign when t is negative, and a positive sign when t is positive, don't I?
     
    Last edited: Feb 6, 2015
  11. Feb 6, 2015 #10

    BvU

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    ##\cos\theta## in post #9 looks good. Easiest to follow with a drawing (xy, d on y-axis). Starts at -1, increases to 0 where it changes sign ( "automatically" ) and keeps increasing to +1. So you see how to draw ##\theta## and which way it is oriented (I had to look several times too... depends on if you take d<0 or d>0 -- the answer for the frequency that the observer hears should not).

    Don't forget to reconstruct f when you are done. The exercise asks for a frequency, not a wavelength :)
    And you can always reverse-engineer the sign so ##f > f_0## for t < 0 and vice versa ;)
     
  12. Feb 7, 2015 #11
    So the expression
    fL= vfs/(v-vscosθ) can be in terms of time if I substitute for cosθ.
    Just considering the denominator for now,
    v-vscosθ = v-vs√(vst2+d2)

    Find a common denominator
    v√(vst2+d2) - vs2t /√(vst2+d2)

    Then subbing that back in
    fL= vfs√(vst2+d2) / v√(vst2+d2) - (vs)2t
     
    Last edited: Feb 7, 2015
  13. Feb 8, 2015 #12

    BvU

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    Aren't you making life difficult ? In the first one I miss a square (v-vs2√...) , in the second I miss the brackets around the numerator and in the and third I miss the brackets around the denominator. Under all squrate roots I miss brackets around vst.
    (Perhaps it's just a typing or formatting issue ? but it's confusing to me...) Fix that and the expressions are correct, but instead of with one square root you end up with two.


    Now we come to the sign. For t<0 we expect fl > fs.

    Top picture is for t<0.
    We want ##v - v_s\cos\theta < v ## for t < 0 so we should write ##v + v_s {v_s t \over \sqrt{v_s^2t^2 + d^2}}## in the denominator !

    upload_2015-2-8_15-15-52.png

    Check with lower picture for t > 0 : we try to have one single expression, so we hang on to ##\cos\theta = - {v_s t \over \sqrt{v_s^2t^2 + d^2}}##. But the angle we want is ## \pi -\theta ## which has the same cosine, but with opposite sign. On the other hand, we want vs negatve, because the source is moving away from the listener.

    So our (awkward?) pick for theta makes that we can use ##v_s \cos\theta = - v_s\; {v_s t \over \sqrt{v_s^2t^2 + d^2}}## once again !

    Conclusion: the same denominator works for T > 0 as well.
    In other words: a plus sign in the denominator of your last expression.

    On other detail: dividing by v (sound velocity) is sensible: vs/v = M, the (dimensionless) Mach number. You could even do it twice and get$$f_l = f_s \;{ 1\over {1+M \;{ t \over \sqrt{t^2+d^2/M^2} } } }
    $$Since you can never do too much checking: for t << 0 that gives ##{1\over 1- M}## and for t >> 0 you find ##{1\over 1 + M}## for the frequency ratio ##f_l/f_s##.

    Lovely !
     

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