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Sound intensity at a microphone

  1. Jan 22, 2008 #1
    1. The problem statement, all variables and given/known data
    A concert loudspeaker suspended high off the ground emits 28.0W of sound power. A small microphone with a 0.900cm^2 area is 51.0m from the speaker.

    A) What is I at the microphone?

    B) How much sound energy impinges on the microphone in J?

    2. Relevant equations
    I = P/A
    I = 4pir^2

    3. The attempt at a solution
    I have solved the answer to part A):

    I = 28W/4pi(51)^2 = 8.57*10^-4 W/m^2


    The answer to part B is what's eluding me:

    I know that at the microphone I = 8.57*10^-4 W/m^2

    So:

    8.57*10^-4 = P/.009m^2

    P = 7.713*10^-6 W which is equal to J/s

    Therefore my answer would be 7.713*10^-6 J of sound energy are impinged each second.

    Mastering Physics is disagreeing with me. Any advice?
     
    Last edited: Jan 22, 2008
  2. jcsd
  3. Jan 22, 2008 #2
    If I am doing something wrong or have not provided enough information, please let me know.

    I must be missing something with this problem, because I can't seem to come up with a different answer than what I have shown.

    Do I have the right approach? Am I making it too complicated by using the Intensity equation twice? Just a small nudge in the right direction would do wonders for me.
     
    Last edited: Jan 22, 2008
  4. Jan 22, 2008 #3
    Okay, took me long enough, but I found my error! As usual, it's a stupid one!

    They reported the area of the mic as .9cm^2, originally I converted the dimension linearly saying that that was equal to .009m^2. In actuality the area is .00009m^2.

    Duh.
     
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