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Doppler Effect?

  1. May 26, 2008 #1
    [SOLVED] Doppler Effect?

    1. The problem statement, all variables and given/known data

    A bat flying at 4.0 m/s emits a chirp at 20 kHz. If this sound pulse is reflected by a wall, what is the frequency of the echo received by the bat?

    2. Relevant equations

    Not sure. Do I use the f(1)=f*[(v+/-observer velocity)/(v+/-source velocity)]

    3. The attempt at a solution

    Not sure where to start. I seem to remember something about sound waves being returned inverted if they hit a wall. Does this factor in?
     
  2. jcsd
  3. May 27, 2008 #2

    alphysicist

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    Homework Helper

    Hi Dreams2Knight,

    As a first step, if the bat emits at a frequency of 20 kHz, what frequency does the wall receive?

    After you have that, the reflection is handled by treating the wall as a source. What do you get?
     
  4. May 27, 2008 #3
    Hello,

    Thank you for replying.

    Source: Bat
    Observer: Wall
    For the first equation I have f(1)=20[345/(345-4)] which is 20.23460411 kHz


    Source: Wall
    Observer: Bat
    For the second equation I have f(1)=20[(345+4)/345] which is 20.23188406 kHz

    Are my equations correct? If so, does the second equation give me my answer, or do I have to do something else?
     
  5. May 27, 2008 #4

    alphysicist

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    The bat emits sound at a frequency at 20 kHz, but the wall does not. After reflection from the wall, what frequency does the sound have? That number is the source frequency f on the right hand side. What do you get?
     
  6. May 27, 2008 #5
    Ok, so if I understand correctly, I need to set my equation up like this:

    20.234=f(2)*[(345+4)/345]. This gives me 20.00209.

    I confused myself at this point because when I put the numbers into my calculator, I accidently subtracted 4 from 345 instead of adding. Doing that gave me the right answer of 20.46 kHz. I was confused because I thought if the source was still, you used V+V(observor) on top, yet subtracting gave me the right answer.

    Then I remembered my teacher mentioning sound waves being inverted when they hit a wall. So, I took the reciprocal of (345+4)/345 and divided 20.234 by that and also got the right answer.

    So, did I have my equation set up correctly as 20.234=f(2)*[(345+4)/345]? Was taking the reciprocal of (345+4)/345 also the correct thing to do?
     
  7. May 27, 2008 #6

    alphysicist

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    No, not quite. It's giving close to the right answer, but I think the way it should be set up is:

    step 1: frequency received by wall:
    [tex]
    f = 20 \frac{345}{345 -4} = 20.2346
    [/tex]

    step 2: frequency received by bat:


    [tex]
    f = (20.2346) \frac{345 + 4}{345} = 20.469
    [/tex]


    Your other ways get close to the answer becuase, for example 1+x is close to 1/(1-x) when x is small compared to one.
     
  8. May 27, 2008 #7

    Ah, I see my mistake now. Thanks so much for your help. I really appreciate it.
     
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