Calculate Distance between A & B: Sound Intensity Problem

[ubiquinone]

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:

A----------B
 \         |
   \       |
     \     |
       \   |
         \ |
           C

 

[Hootenanny]

Have you any thoughts yourself?

 

[ubiquinone]

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!

 

[Hootenanny]

Perhaps this ; http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.html#c1 would be good background reading.

 

[ubiquinone]

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 $$\displaystyle\frac{I_1}{I_2}=\frac{r_2^2}{r_1^2}\Rightarrow \frac{10^{-7}}{10^{-7.47}}=\frac{\sqrt{x^2+10^2}^2}{x^2}$$
$$\displaystyle\Leftrightarrow 10^{0.47}x^2=x^2+100$$
$$\displaysytle\Leftrightarrow x^2(-1+10^{0.47})=100$$
$$\displaystyle\Rightarrow x=\sqrt{\frac{100}{10^{0.47}-1}}\approx 7.16m$$