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Calculate the decibel level

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    A car horn sounds and 15 m away the sound level is recorded as 85 dB.
    Calculate the decibel level 25 m away from the horn.

    2. Relevant equations

    β = (10dB)logI/Io

    β = (20dB)logP/Po

    3. The attempt at a solution

    I = P/A @ 15m
    A = 4∏r2
    A = 2827
    I = 15/2827 = 0.005W/m

    β = 10dBlog0.005/10-12
    = 87dB

    I = P/A @ 25m
    A = 4∏r2
    A = 7854
    I = 25/7854 = 0.003W/m

    β = 10dBlog0.003/10-12
    = 85dB

    I dont know if I'm even heading in the right direction with this one!
     
  2. jcsd
  3. Jan 24, 2012 #2

    I like Serena

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    Hi ninaw21! :smile:

    How did you get that P would be 15? It isn't true. P is unknown (as yet).

    Please note that you are using two different versions of P here.
    The P in your relevant equation is actually the sound pressure, usually denoted with a small p.
    The P in the equation I = P/A is the power of the source.


    You do have that at 15 m: β = (10dB)logI15/Io = 85 dB
    and furthermore that I15 = P/A at 15 m.
    From this you can calculate what I25/I15 is. Do you know how?
     
  4. Jan 24, 2012 #3
    Thank you! No I dont know how to calculate it can you explain it please?
     
  5. Jan 24, 2012 #4

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    You already wrote that A = 4∏r2.

    So I15 = P / 4∏152.
    And I25 = P / 4∏252.
    In both cases P is the (unknown) power of the source.

    Can you divide I25 by I15?
     
  6. Jan 24, 2012 #5
    I divided I25 by I15 and got a decibel level of 3dB..Is that not very low?
     
  7. Jan 24, 2012 #6
    do i add it to the original decibel level?
     
  8. Jan 24, 2012 #7

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    Hmm, suppose you add 3dB to the original decibel level... then the sound would be louder at 25 m? That can't be right...

    Anyway, if I try to find the dB level, I do not get 3 dB...

    Let's do one thing at a time.
    What is I25/I15?

    After that, what is the corresponding decibel level, which is 10log(I25/I15)?
     
  9. Jan 24, 2012 #8
    I25/I15 = 2.78

    10log2.78 = 4dB
     
  10. Jan 24, 2012 #9

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    Let's see...

    I25 / I15 = (P / 4∏252) / (P / 4∏152) = 152 / 252.

    erm... no that is not 2.78...
     
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