1. Not finding help here? Sign up for a free 30min 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!

Simple sound intensity difficulty

  1. Nov 9, 2007 #1
    According to the text, at a distance of 50 m from a jet engine, the sound intensity is 10 W/m2. This corresponds to a loudness (or an intensity level) of 130 dB. [You should convince yourself that this is true.] If you find the sound too loud and you want to reduce the intensity by a factor of 10, to what new distance (approximately) from the jet engine should you move to achieve this reduction?


    I=P/Area

    Intensity level = 10db*log(I/Io)


    to find the power of the jet engine:
    p=10*((50^2)*4pi)
    =314159

    a factor of 10 reduction would mean 13dB
    13=10log(I/10^-12)
    1.3=log(I)-log(10^-12)
    -10.7=log(I)
    I=10^-10.7

    A=P/I
    A= 314159/ 10^-10.7
    A=1.57x10^16

    1.57x10^16 = 4pi r^2
    r=397289

    so does that mean that you really can't make the intensity decrease by a factor of 10?
     
    Last edited: Nov 9, 2007
  2. jcsd
  3. Nov 9, 2007 #2
    I don't think dB need to come into the calculation; the question asks for a reduction in intensity, not in "intensity level" or loudness.

    The requisite new level is a factor of 10 less than the first level of 10 W/m^-2, that is 1 W/m^-2.

    Assuming the jet engine is on a plane surface and there are no reflections then the noise power is radiating through the surface of a hemisphere ...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Simple sound intensity difficulty
  1. Sound intensities (Replies: 1)

  2. Intensity of sound (Replies: 5)

Loading...