Calculate Distance between A & B: Sound Intensity Problem

AI Thread Summary
The discussion revolves around solving a sound intensity problem involving distances between points A, B, and C. A listener at point B measures sound intensity at 50.0 dB, while at point C, it is 45.3 dB, with a known distance of 10.0 m between B and C. The user seeks assistance in applying physics concepts to calculate the distance from A to B, denoted as \overline{AB}. Through calculations involving sound intensity formulas and ratios, the solution reveals that the distance \overline{AB} is approximately 7.16 m. The discussion emphasizes the application of theoretical physics to practical problem-solving.
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Hi I have a question here from a chapter on sound. I'm not sure on how to solve this, so I was wondering if someone here could please give me a hand. Thank You!

Question: A point source at A emits sound uniformly in all directions. At point B, the listener measures the sound intensity to be 50.0dB. At point C, the sound intensity level is 45.3dB. The distance \overline{BC} is 10.0m. Calculate the distance \overline{AB}.

Diagram:
Code:
A----------B
 \         |
   \       |
     \     |
       \   |
         \ |
           C
 
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Have you any thoughts yourself?
 
This question is from an old physics exercise book which does not have any examples. I have tried solving it by reading up the concepts on waves and sound from a conceptual physics book. I'm hoping if someone here can show me how to solve this problem, so I can see and understand how a physicist or a good problem solver applies physics concepts and theory into problem solving. Again, any help with this would be greatly appreciated. Thanks!
 
Hi Hootenanny, thank you for the reference! I think I got it now.

Let x be the distance of \overline{AB}
Calculate the intensities:
\displaystyle 50.0dB=10\log\left (\frac{I_1}{10^{-12}}\right )
\displaystyle\Leftrightarrow 10^5=\frac{I_1}{10^-12}
\displaystyle\Leftrightarrow 10^{-7}=I_1
\displaystyle 45.3dB=10\log\left (\frac{I_2}{10^{-12}}\right )
\displaystyle\Leftrightarrow 10^{4.53}=\frac{I_2}{10^-12}
\displaystyle\Leftrightarrow 10^{-7.47}=I_2
Since \displaystyle\frac{I_1}{I_2}=\frac{r_2^2}{r_1^2}\Rightarrow \frac{10^{-7}}{10^{-7.47}}=\frac{\sqrt{x^2+10^2}^2}{x^2}
\displaystyle\Leftrightarrow 10^{0.47}x^2=x^2+100
\displaysytle\Leftrightarrow x^2(-1+10^{0.47})=100
\displaystyle\Rightarrow x=\sqrt{\frac{100}{10^{0.47}-1}}\approx 7.16m
 
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