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Thread: 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): $\displaystyle L = 10 \log (\frac{I}{10^{-12}})$

    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 $\displaystyle 60dB$:

    $\displaystyle
    60 = 10 \log (\frac{I}{I{o}})$
    $\displaystyle
    6 = log(\frac{I}{I{o}})$
    $\displaystyle
    10^6 = (\frac{I}{I{o}})$
    $\displaystyle
    10^6{I{o}} = I
    $

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

    $\displaystyle
    dB = 10 \log (\frac{2I}{I{o}})$

    Since $\displaystyle I = 10^6{I{o}} $:

    $\displaystyle
    dB = 10 \log (\frac{2{10^6{I{o}}}}{I{o}})$
    $\displaystyle
    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***

    $\displaystyle
    dB = 10 (2*6)$
    $\displaystyle
    dB = 10(12)$
    $\displaystyle
    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 $\displaystyle 60dB$:

    $\displaystyle
    60 = 10 \log (\frac{I}{I{o}})$
    $\displaystyle
    6 = log(\frac{I}{I{o}})$
    $\displaystyle
    10^6 = (\frac{I}{I{o}})$
    $\displaystyle
    10^6{I{o}} = I
    $

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

    $\displaystyle
    dB = 10 \log (\frac{2I}{I{o}})$

    Since $\displaystyle I = 10^6{I{o}} $:

    $\displaystyle
    dB = 10 \log (\frac{2{10^6{I{o}}}}{I{o}})$
    $\displaystyle
    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***

    $\displaystyle
    dB = 10 (2*6)$
    $\displaystyle
    dB = 10(12)$
    $\displaystyle
    dB = 120 $

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