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Figuring out changes in Intensity (Inverse Square Law)

  1. Jan 12, 2015 #1
    1. The problem statement, all variables and given/known data

    If I measure a sound intensity of 1.0 at distance R from its source, what intensity would I measure at distance 3R in a free, unbounded space? What is the difference in decibels?

    2. Relevant equations

    I know that the equation for this is 10 (log R2/R1)

    3. The attempt at a solution

    I thought that the answer could be a 9.54 dB difference. I did: 10(log(9) to figure this out but the answer doesn't seem correct. Any help with answering this question would be greatly appreciated. Thank you
     
  2. jcsd
  3. Jan 12, 2015 #2

    gneill

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    Staff: Mentor

    Intensity follows an inverse square law with distance. So you'd expect the intensity at 3R to be 9 times less than the intensity at R, right? So the inverse square law for intensity vs distance looks like:
    $$\frac{I_2}{I_1} = \frac{R_1^2}{R_2^2}$$
    Decibels are a comparison of intensities, so if you want to use the distances in the decibel formula rather than the intensities you need to remember to keep the inverse square law in mind and employ the squares of the distances and the "inverse" property as well. That said, take a look at the ratio above and then your Relevant equation. Does your equation preserve the "inverse" and "square" properties?
     
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