# Intensity of sound wave question

The question is:

The intensity of sound wave A is 100 times weaker than that of sound wave B. Relative to wave B the sound level of wave A is?

I tried doing (10dB)Log(1/100) but that equals -20dB

thanks

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The answer of -2 dB is wrong. -2 dB is an extremely small difference! You can verify in Wikipedia's article on decibels that a power ratio of 100 corresponds to a difference of 20 dB. –  Rahul Jun 7 '12 at 8:05

Your answer of $-20$ dB is correct.

I suspect that you’re using a definition that gives the sound level in decibels as $10\log\left(\frac{I}{I_0}\right)$, where $I_0$ is the minimum perceptible intensity, or something very similar; if so, the following calculation shows exactly why $-20$ dB is correct.

Suppose that $I_A$ is the intensity of sound wave $A$, and $I_B$ is the intensity of sound wave $B$. Then the loudness of $A$ in decibels is $$L_A=10\log\left(\frac{I_A}{I_0}\right)\;,\tag{1}$$ and that of $B$ is $$L_B=10\log\left(\frac{I_B}{I_0}\right)\;.$$

You’re told that $I_A=\frac1{100}I_B$. Substituting that into $(1)$, we get

\begin{align*} L_A&=10\log\left(\frac{I_A}{I_0}\right)\\ &=10\log\left(\frac{\frac1{100}I_B}{I_0}\right)\\ &=10\log\left(\frac{\frac{I_B}{I_0}}{100}\right)\\ &=10\left(\log\left(\frac{I_B}{I_0}\right)-\log 100\right)\\ &=10\log\left(\frac{I_B}{I_0}\right)-10\cdot2\\ &=L_B-20\;. \end{align*}

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Thank you man. Very clear explanation. –  Petersicecream Jun 7 '12 at 8:28

Just 10*log10(1/100) = -20 dB and you are done! @Petersicecream: you are right, multiply by 10. Thanks!

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don't you have to multiply it by 10dB? from the equation? –  Petersicecream Jun 7 '12 at 7:54
@Petersicecream: Yes, you’re right. Sound $A$ is $2$ bels less loud than sound $B$, but that makes it $20$ decibels less loud. –  Brian M. Scott Jun 7 '12 at 8:11