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Sound intensity level

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data

    A sound wave is measured to have a 20-decibel sound intensity level (SIL). And the sound wave power flux (measured in Watts/m2) is then increased by a factor of 2000. What is the new sound level (in decibels) after this factor of 2000 increase in sound power?

    2. Relevant equations

    Sound intensity level= 10log(I/I_0)

    3. The attempt at a solution

    10log(20 / 10-12W/m2) = 10log(20000000 MegaWatt) = 7.301

    One thing i do not understand is this factor increase by 2000, therefore i am not sure what to do next.

    Thanks for your help in advance
  2. jcsd
  3. Aug 24, 2010 #2


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    Take a look at your relevant equation. I, in the equation, is not 20 dB. Instead, I is is something measured in W/m2. The left side of the equation, the "Sound intensity level," is what is 20 dB. What happens to this "Sound intensity level (in dB)" (which is on the left side of the equation) when I is increased by 2000? :wink:
  4. Aug 24, 2010 #3
    Firs of all, thank you for your willingness to help.

    then 20dB = 10log(I+2000/10-12W/m2)

    is that correct? and what is I?

    Sorry for my stupidity in physics:}
  5. Aug 24, 2010 #4
    I think i know now:

    I = 0 + 2000 Wm-2?

    If yes, then

    20dB = 10log(2000 Wm-2 /10-12W m2)
    20dB = 10log(2000 / 0.0000000000001W m2)
    20dB = 10log(20000000000000000)
    20dB = 12.529

    Somethings not right:/
    Last edited: Aug 24, 2010
  6. Aug 24, 2010 #5


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    Not quite. The new I is the old I multiplied by 2000 (not added).

    So originally we have

    [tex] 20 \ \mbox{[dB]} = 10 \log \left( \frac{I}{I_0} \right) [/tex]

    Find the new "Sound intensity level"

    [tex] \mbox{Sount intestiy level [dB]} = 10 \log \left( (2000) \frac{I}{I_0} \right) [/tex]
  7. Aug 24, 2010 #6
    i get it now, but then i don't understand the value of I (?)
  8. Aug 24, 2010 #7


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    In the equations we've used above, I is the power flux, as measured in units of W/m2.

    Here, power flux is really almost the same thing as sound intensity, conceptually anyway. The sound intensity is a unit-less ratio converted to "units" of dB. But since the ratio is always relative to a constant, unchanging value, the sound intensity is just another way to represent the power flux. They are kind of the same thing, just represented differently with different units.

    And units of dB are barely units. Units of "dB" are not units in the normal sense. The "dB" values describes the scaling of a unit-less power ratio. And value expressed in units of dB is either representing some sort of scaling (i.e. multiplication) factor, or by how much power something has relative to something else. (Note: there is an inherent assumption though that the ratio in the [itex] 10 \log (I/I_0) [/tex] formula is a ratio of powers, power fluxes, or something having to do with power)

    Going back to your specific problem, if it makes it any easier to visualize, after the power flux is increased by a factor of 2000, the equation is

    \mbox{Sount intestiy level [dB]} = 10 \log \left(\frac{2000 \times I}{I_0} \right)

    If you'd rather say that I2 is the power flux after it gets increased by a factor of 2000 (in other words, I2 = 2000 x I), then the equation is

    \mbox{Sount intestiy level [dB]} = 10 \log \left(\frac{I_2}{I_0} \right)
    Last edited: Aug 25, 2010
  9. Aug 24, 2010 #8
    ;/ damn it, i still do not understand the last line. Like should i convert I2 and I0 to digits or what... Looks like it's still too hard for me to handle this level.

    I don't think i will be able to finish this on my own.

    But thank you for your explanations, i really appreciate it.
  10. Aug 24, 2010 #9


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    My last post was merely an attempt to explain the relationship between variables in the equation. It wasn't really meant to help you solve the particular problem.

    But if you haven't solved the problem yet, there are a couple of ways to do it. One way is to solve for I in your first equation (where you know the sound intensity level is 20 dB.) Multiply that I by 2000, and solve for the new sound intensity level.

    A different way to do it is to recognize that

    [tex] \log(ab) = \log(a) + \log(b) [/tex]
  11. Aug 24, 2010 #10
    Intensity level = 10log(2000x20 / 0.0000000000001)
    Intensity level = 10log(400000000000000000)
    Intensity level = 17.602dB.

    Is this correct?
  12. Aug 25, 2010 #11


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    Sorry, not quite correct. :cry:

    I is not equal to 20 dB. I is in units of W/m2, not dB.

    Don't mix up the sound intensity level with the power flux. Like I said before, they are different expressions for the same thing, but they are not the same number.

    Let me give you a start, doing it the long way (you can solve the problem the easier way too, but I think it's important to understand how to do it this way, at least in concept).

    [tex] 20 \ [\mbox{dB}] = 10 \log \left( \frac{I}{1 \times 10^{-12} \left[ \frac{\mbox{W}}{\mbox{m}^2} \right]}\right) [/tex]

    Divide both sides of the equation by 10.

    [tex] 2 = \log \left( \frac{I}{1 \times 10^{-12} \left[ \frac{\mbox{W}}{\mbox{m}^2} \right]}\right) [/tex]

    Bring each side to the power of 10.

    [tex] 10^2 = 10^{\log \left( \frac{I}{1 \times 10^{-12} \left[ \frac{\mbox{W}}{\mbox{m}^2} \right]}\right)}[/tex]

    Simplify the right side,

    [tex] 10^2 = \frac{I}{1 \times 10^{-12} \left[ \frac{\mbox{W}}{\mbox{m}^2} \right]} [/tex]

    Multiply each side by 1 x 10-12 [W/m2]

    [tex] (10^2) \left(1 \times 10^{-12} \left[\frac{\mbox{W}}{\mbox{m}^2} \right] \right) = I [/tex]

    Simplify the left side,

    [tex] 1 \times 10^{-10} \left[ \frac{\mbox{W}}{\mbox{m}^2} \right] = I [/tex]

    Now multiply I times 2000 and convert to dB.

    Alternately, Like I mentioned before, there is an easier way to do this problem, where you don't even need to even know that I0 is 1 x 10-12 [W/m2]. You should do it that way too to double check your answer. (Hint: Use the clue I gave in post# 9.)
    Last edited: Aug 25, 2010
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