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Sound Intensity problem. Need some help.

  1. Apr 20, 2004 #1
    I have no idea what I'm doing wrong on this one. I'm having problems somewhere in the log conversions to W/m^2 I believe.

    A siren mounted on the top of towers produces a noise of 120 dB at a distance of 2 meters. Treating the siren as a point particle (and ignoreing reflection and absorbtion) find the intensity level heard by an observer (a) 12 m away (b) 21 m away. (c) How far away can the sound be heard?

    First I thought this was going to be an easy plug in the formula problem. Where I2=(r1/r2)^2 *I1
    I think I need to convert this to W/m^2, and then plug it into this equation, and then convert back. However, I'm not quite sure how to do this. I think it has something to do with Β=10 log (I/Io) Where Io is 10^-12 W/m^2

    So we have B=1200 (thats 120 dB times 10)
    120=10 Log (I/(10^-12)) Can you help me solve this one? Thanks a million!
     
  2. jcsd
  3. Apr 20, 2004 #2

    Doc Al

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    Staff: Mentor

    Hint: By what ratio is the intensity decreased at 12m from the source compared to 2m? What is that intensity ratio in db's? Remember that logarithms add (or subtract).
     
  4. Apr 20, 2004 #3
    I appreciate the advice, but I can't seem to figure it out with that. However, I begin to think that if 10 Log(10)=1 and 10 Log(100)=2 Wouldnt the equation of (I/Io) equal 1 with 108 zeros? (thats +120 because of the given dB and -12 because youd be multipling that by 10^-12)?
    Then you can multiply this number (too long for my ti-83) by (2/12)^2 and plug back into log. But alas, this methods fails (although it gives a reasonable answer much closer than my previous method). I got 107 Db where it should have been 104. Part(B) is about the same thing where it should be 99.6. I'd be willing to bet I'm making this a lot harder than it is. Any additional advice for the man who conceptually digs his own grave?
     
    Last edited: Apr 20, 2004
  5. Apr 20, 2004 #4

    Doc Al

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    Of course you are. :smile:

    Try this. Imagine three different intensities: I0, A, and B. If A is 100 times I0, what is it's intensity in db? = 10 log (100) = 20 db, right? What if B is 10 times the intensity of A? That intensity difference, in db, equals 10 log(B/A) = 10 log(10) = 10 db. So... what's the intensity of B with respect to I0? It's just 20db + 10db = 30 db. Got it?

    Check that with the basic definition: Intensity B in db = 10 log(B/I0) = 10 log (1000) = 30 db.

    Let's apply this to your first problem: intensity at 12 m compared to 2 m. I assume an inverse square law, so the intensity at 12 m equals 1/36 of that at 2 m. So... 10 log (1/36) = - 15.5 db. Thus, intensity at 12 m = 120 db -15.5 db = 104.5 db.
     
  6. Apr 20, 2004 #5
    Allright. I understand what your saying but I have a couple questions. First of all isn't 10Log(100) equal to 2 (I think log is naturally set to 10Log in most calculators). When I type 10Log(100) into my ti-83 it gives me 20. But when I type Log[10,100] into Mathematica it gives me 2. I guess since you work checked out answer wise, this means the equation itself ALSO assumes the 10Log?
    And second, you mentioned an inverse square law. While this isn't in my book it very well may have been covered the class I missed (hence more trouble than usual with this problem). Did you just take the ratio of distance (12m/2m) square this and then inverse it? If so this process should be the same for all similar problems shouldn't it? Thanks so much for helping me as much as you do. Haha. I should send you some money over paypal sometime. :biggrin:
     
    Last edited: Apr 20, 2004
  7. Apr 20, 2004 #6

    Doc Al

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    Staff: Mentor

    Hmmm... maybe this is your problem...

    10 log (100) means 10 x log(100) = 10 x 2 = 20. (I assume by log, we both mean log base 10.) So I presume that's what you typed into your calculator, and that's what you got. In Mathematica, you found the log base 10 of 100, which equals 2. You'd better learn how to use these tools!

    Do you really need a calculator to find 10 log (100)? :rolleyes:
    Yes, that's the idea of the inverse square law. It should apply to any point source, unless otherwise noted. In your first post you wrote:
    that's the inverse square law.
     
  8. Apr 20, 2004 #7
    Click! Thank You So Much!!!! Your right I do need to learn how to use these tools a little better. And haha no, I can take care of 10Log(100) in my head, providing I would have naturally assumed the whole 10(10Log) thing (haha what a coincidence). Thanks again!
     
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