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Math Help - Sound Level Intensity (Logarithm)

  1. #1
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    Sound Level Intensity (Logarithm)

    Hey Guys, I have a simple question about sound level intensity (Log.)... so here it goes.

    2 friends are walking and having normal conversation; the sound level of the conversation is 60 dB . When they get near a busy street, they double their intensity of the sound to be able to hear each other. What is the new sound level ?

    Can some please help me with this question?...
    Thanks

    PS: You cannot just double the sound level intensity..
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  2. #2
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    Quote Originally Posted by MordernWar2 View Post
    Hey Guys, I have a simple question about sound level intensity (Log.)... so here it goes.

    2 friends are walking and having normal conversation; the sound level of the conversation is 60 dB . When they get near a busy street, they double their intensity of the sound to be able to hear each other. What is the new sound level ?

    Can some please help me with this question?...
    Thanks

    PS: You cannot just double the sound level intensity..

    Sound Level (dB): L = 10 \log (\frac{I}{10^{-12}})

    Are you not able to use the formula to see what happens when you double intensity
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  3. #3
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    Since a normal conversation is 60dB:

    <br />
60 = 10 \log (\frac{I}{I{o}})
    <br />
6 = log(\frac{I}{I{o}})
    <br />
10^6 = (\frac{I}{I{o}})
    <br />
10^6{I{o}} = I<br />

    Now that questions states that the intensity doubled near a busy street:

    <br />
dB = 10 \log (\frac{2I}{I{o}})

    Since  I = 10^6{I{o}} :

    <br />
dB = 10 \log (\frac{2{10^6{I{o}}}}{I{o}})
    <br />
dB = 10 \log (2*10^6)

    ***I'm not sure if the next step is correct, though essentially you would cancel out log and base 10***

    <br />
dB = 10 (2*6)
    <br />
dB = 10(12)
    <br />
dB = 120

    Hope that helps!
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  4. #4
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    Quote Originally Posted by Sigra View Post
    Since a normal conversation is 60dB:

    <br />
60 = 10 \log (\frac{I}{I{o}})
    <br />
6 = log(\frac{I}{I{o}})
    <br />
10^6 = (\frac{I}{I{o}})
    <br />
10^6{I{o}} = I<br />

    Now that questions states that the intensity doubled near a busy street:

    <br />
dB = 10 \log (\frac{2I}{I{o}})

    Since  I = 10^6{I{o}} :

    <br />
dB = 10 \log (\frac{2{10^6{I{o}}}}{I{o}})
    <br />
dB = 10 \log (2*10^6)

    ***I'm not sure if the next step is correct, though essentially you would cancel out log and base 10***

    <br />
dB = 10 (2*6)
    <br />
dB = 10(12)
    <br />
dB = 120

    Hope that helps!
    No, that last bit isn't correct. Instead evaluate dB = 10 \log (2*10^6) directly
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