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Intensity of a Loudspeaker, What's wrong with calculation?

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data
    A certain loudspeaker system emits sound isotropically with a frequency of 1770 Hz and an intensity of 0.892 mW/m2 at a distance of 5.66 m. Assume that there are no reflections. (a) What is the intensity (in mW/m2) at 30.8 m? At 5.66 m, what are (b) the displacement amplitude and (c) the pressure amplitude of the sound? Take the speed of sound to be 343 m/s and the density of air to be 1.21 kg/m3.

    I mostly need help with part b, I believe I can get part c with the answer from part b.

    2. Relevant equations

    I = 1/2(ρvω2sm2)

    3. The attempt at a solution

    So, for part b and from the relevant equation:

    sm = (2I/ρvω2)1/2

    ω = 2πf = 2*3.14159*1770 = 11121.24 Hz

    So, using the above equation and plugging in I = 0.892 mW/m2, ρ = 1.21 kg/m2, and v = 343 m/s, I get the answer 5.895 x 10-6 m. However, this answer is apparently wrong, and I'm not sure why. Is there something wrong with my calculations?
     
    Last edited: Sep 5, 2011
  2. jcsd
  3. Sep 5, 2011 #2
    intensity is the power/area covered. Such areas are usually spherical. so at a distance of 5.66meters,

    .892E6W=
    8.92E5W/m^2=power(dE/dt)/4pie(5.66)^2 if this really is a sphere and get power using algebra.

    once you get power, intesity=power/4pie(30.8)^2 I dont get neither b or c. lol I didnt study pressure yet. I don't even know what amplitude affects.
     
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