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Sound Waves

  • Thread starter ubiquinone
  • Start date
43
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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 [tex]A[/tex] emits sound uniformly in all directions. At point [tex]B[/tex], the listener measures the sound intensity to be [tex]50.0dB[/tex]. At point [tex]C[/tex], the sound intensity level is [tex]45.3dB[/tex]. The distance [tex]\overline{BC}[/tex] is [tex]10.0m[/tex]. Calculate the distance [tex]\overline{AB}[/tex].

Diagram:
Code:
A----------B
 \         |
   \       |
     \     |
       \   |
         \ |
           C
 

Hootenanny

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Have you any thoughts yourself?
 
43
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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!
 
43
0
Hi Hootenanny, thank you for the reference! I think I got it now.

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

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