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Find the speed of the source (Doppler Effect)

  1. Sep 13, 2015 #1
    A kid is riding a bike and is ringing bell of 3000 Hz. You are standing still on the sidewalk.When the bicycle moves toward you, you hear a shift in the frequency of the bell. In addition, when the bicycle moves away from you, you hear a different shift in the frequency of the bell. If there is a 30 Hz change in the frequencies that you hear for the bell as the bicycle approaches you and then moves away from you, what is the speed (in m/s) of the bicycle? (Write your answer to the nearest 0.01 m/s.)

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

    Frequency(source)(fs) = 3000 Hz
    speed(observer)(vo) = 0 m/s
    Frequency(towards)(ft)= 3030 Hz (NOT SURE)
    Frequency(away)(fa) = 2970 Hz (NOT SURE)
    speed of sound(v) = 343 m/s

    2. Relevant equations

    fo = ((v+vo)/(v-vs)) * fs Bike is moving TOWARDS me
    fo = ((v-vo)/(v+vs)) * fs Bike is moving AWAY from me

    3. The attempt at a solution

    Towards
    ------------
    3030Hz/3000Hz = (343 m/s) / (343 m/s - vs)
    (1.01)(343 m/s - vs) = 343 m/s
    345.20 m/s - 1.01 vs = 343 m/s
    vs = 2.18 m/s

    I do the same thing using the AWAY equation, and i get the same answer.
    However, that answer is incorrect. I do not know what I am doing wrong.
     
  2. jcsd
  3. Sep 13, 2015 #2

    mfb

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    The 30 Hz difference are between "towards you" and "away from you". They are not relative to the 3000 Hz.
     
  4. Sep 13, 2015 #3
    Could you please elaborate on that. And, give me a hint.
     
  5. Sep 13, 2015 #4

    mfb

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    If the bike approaches you, the frequency is x, a bit higher than 3000 Hz.
    If the bike moves away from you, the frequency is y, a bit lower than 3000 Hz.
    x - y = 30 Hz.
    Find formulas for x and y and solve the equations, there is no trick involved.
     
  6. Sep 13, 2015 #5
    That's what I am doing. I added and subtracted 30Hz.
     
  7. Sep 13, 2015 #6

    mfb

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    You used x - 3000 Hz = 30 Hz and 3000 Hz - y = 30 Hz, but that's not right. It would imply x-y = 60 Hz.
     
  8. Sep 13, 2015 #7
    So you are saying that 30Hz is the total change in frequency I hear for towards and away. That means that I hear 15 Hz for towards and 15 Hz for away.
    If so, then it is 3000-15=2985(away) and 3000+15=3015(towards). Then I just use the equation I used above to get the answer.
     
  9. Sep 13, 2015 #8

    mfb

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    That's what the problem statement says.
    It is not, the doppler shift is not symmetric.
     
  10. Sep 13, 2015 #9
    I don't think I get it yet. You will need to help me with equations.
     
  11. Sep 13, 2015 #10

    mfb

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    All necessary equations are in the thread already, you just have to combine them.
     
  12. Sep 13, 2015 #11
    You know what I am trying to tell you. If I had understood what you were trying to tell me, I would have done it by now.
    I don't get it.
     
  13. Sep 13, 2015 #12
     
  14. Sep 13, 2015 #13

    mfb

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    No I don't know. The problem got reduced from a problem about the Doppler effect to a topic your classes covered several years ago: plugging in equations and numbers into each other.
     
  15. Sep 13, 2015 #14
    You have been telling me to do the same thing over and over, and I don't understand it. You are an expert. And, I appreciate your help.
    But, if I am your student in this case, and I am not understanding it, then you should help me with a different approach. Please, help me in a different way.
     
  16. Sep 13, 2015 #15

    mfb

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    There is only one reasonable approach, and I don't know how to explain it differently, sorry.
     
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