1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sound Waves

  1. Jan 7, 2007 #1
    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].

    Code (Text):

     \         |
       \       |
         \     |
           \   |
             \ |
  2. jcsd
  3. Jan 7, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Have you any thoughts yourself?
  4. Jan 7, 2007 #3
    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!
  5. Jan 7, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  6. Jan 8, 2007 #5
    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]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook