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Doppler effect shifts

  1. Dec 6, 2016 #1
    1. The problem statement, all variables and given/known data

    An audio transmitter and receiver are mounted side by side as shown in the figure. The transmitter emits sound of frequency v. A distant flat plate approaches the instrument at speed Vtarget and the receiver detects sound waves of frequency v' reflected from the target. The difference in frequency between the emitted and detected frequencies can be used as a motion detector.

    A) Show that v'/v=v+Vtarget / v-Vtarget

    B) In many cases V[/SUB]target[/SUB] <<v . For such cases show that V'/v=1+2(vtarget /v)

    FIGURE: http://imgur.com/bLXnm2L
    2. Relevant equations
    Doppler effect

    3. The attempt at a solution
    Ive already accomplished A by realizing there were two doppler shifts and that the frequency gets larger as it goes to the object, than even larger than that when it bounces back to the receiver. Part B is where i'm stuck because im not sure what i need to assume to create that formula.
     
  2. jcsd
  3. Dec 6, 2016 #2

    haruspex

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    Do you know how to expand (1+x)a as a power series in x?
     
  4. Dec 6, 2016 #3
    No, i havent learned about power series yet, how does that relate to this?
     
  5. Dec 7, 2016 #4

    haruspex

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    For this question, you need to know that, for small x, (1+x)a can be approximated as 1+ax+O(x2), where O(x2) means terms that grow smaller "like x2" as x gets smaller. I.e. you can just approximate here as 1+ax.
     
  6. Dec 7, 2016 #5
    is x just a term for frequency in this situation? Im not sure im following your variables.
     
  7. Dec 7, 2016 #6

    haruspex

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    x is anything at all, as long as it is small compared with 1.
    Your initial post is very confusing because you seem to have written v for frequency and for velocity of sound in air. Maybe it's just a confusing font. Allow me to rewrite it more clearly:
    You are given Vtarget<<Vsound. So what quantity << 1?
     
  8. Dec 7, 2016 #7
    so your variable x would be less than 1, or in the re formatted variable, Vsound would be <<1
     
  9. Dec 7, 2016 #8

    haruspex

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    Yes.
    No, Vtarget<<Vsound. From that, find something that must therefore be <<1.
     
  10. Dec 7, 2016 #9
    Perhaps the frequency would be less than 1. Would i be correct in assuming that?
     
  11. Dec 7, 2016 #10

    haruspex

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    No.
     
  12. Dec 7, 2016 #11
    OH. The Vtarget has to be <<1
     
  13. Dec 7, 2016 #12

    haruspex

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    No.
    Forget the actual questionfor a moment. If you have two quantities y and z, and you are told y<<z, what quantity can you construct from y and z which is therefore << 1? (You understand what << means, right?)
     
  14. Dec 7, 2016 #13
    << means significantly less than. So id have to make a third quantity that is z-y?
     
  15. Dec 7, 2016 #14

    haruspex

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    Yes.
    No. E.g. y=1 and z=100 would satisfy y<<z, but z-y would be 99. And I should have said that you want the absolute magnitude to be <<1, so swapping z and y won't do it either.
    Think about ratios.
     
  16. Dec 7, 2016 #15
    So what if we made a quantity that was 1/y. that would guarentee y<<Z if we just want it to be <<1
     
  17. Dec 7, 2016 #16
    whoops, that wouldnt work either because if y was 1/3 than it would be 1/(1/3) which is >>1. ill keep thinking.
     
  18. Dec 7, 2016 #17
    nvm im confused again
     
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