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Isotropic loudspeaker - calc based

  1. Nov 10, 2006 #1
    1) Isotropic loudspeaker A certain loudspeaker system emits sound isotropically
    with a frequency of 2.00 x 103 Hz and an intensity of 1.00 x 10-3 W/m2 at a distance of 7.00 m.
    Assume there are no reflections. Use 344 m/s for the velocity of sound in air and 1.21 kg/m3 for
    the density of air.
    a) What is the displacement amplitude at 7.00 m?
    b) What is the pressure amplitude at 7.00 m?
    c) What is the intensity at 25.0 m?

    Hi, does the term isotropic change the way I calculate the displacement amplitude or does the wave still propagate in the +x direction?

    I want to use these formulas,

    v = sqrt(B/p) B is the bulk modulus and p is the density.
    v = (f)(wavelength)

    I = 1/2(sqrt(pB)(w^2)(A^2) = (p_max^2)/(2pv)

    = (p_max^2)/(2(sqrt(pB))

    for the intensity formula denoted I. p = density and p_max is the pressure amplitude which is directly proporional to A, the displacement amplitude.

    please let me know if I can use these formulas for this case.. thank you!!
  2. jcsd
  3. Nov 10, 2006 #2

    Doc Al

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    Staff: Mentor

    Isotropic just means that the sounds travels equally in all directions, so the intensity will drop off as 1/r^2.

    Those equations are OK, but you should be able to write the intensity as a function of pressure amplitude or displacement amplitude without needing the bulk modulus. (Since you have density and speed as given.)
  4. Nov 10, 2006 #3
    do you mean something like this

    p_max(x) = sqrt((I)(2pv))sin(kx)
    Last edited: Nov 10, 2006
  5. Nov 10, 2006 #4

    Doc Al

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    Staff: Mentor

    No. Start with your equation:
    The right hand term gives you intensity as a function of pressure amplitude.

    To find intensity as a function of displacement amplitude, eliminate "sqrt(pB)" from the middle term.

    Neither equation will have B.
  6. Nov 10, 2006 #5

    a ) 1.00 x 10-3 W/m2 = (p_max^2)/(2(1.21 kg/m3)(344m/s)
    (p_max) = sqrt((1.00 x 10-3 W/m2)(2(1.21 kg/m3)(344m/s))
    = .91m

    b) then the equation just becomes I = (w^2)(A^2) = y_max?

    w = vk
    k = 2pi/wavelength
    v = (f)wavelenth
  7. Nov 10, 2006 #6

    Doc Al

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    Staff: Mentor

    That's pressure amplitude, not displacement amplitude. (Units!)

    Huh? y_max is A! You already had the equation:
    Just replace sqrt(pB) to eliminate B from the expression. (Hint: v = sqrt(B/p)... so what's sqrt(pB)?)
  8. Nov 10, 2006 #7

    I = 1/2pv(w^2)(A^2)


    thank you!
  9. Nov 10, 2006 #8

    Doc Al

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    Staff: Mentor

    That's the one. (Relate w to frequency and you're done. Then you can solve for A.)
  10. Nov 10, 2006 #9
    w = vk
    k = 2pi/wavelength
    v = (f)wavelenth

    ok i just want to make sure that this is right,

    v/f = wavelength
    344/2.00 x 10^3 = .172m

    so k = 2pi/.172 = 36.53

    w = vk = 344 * 36.53 = 12566.3

    so (A) = sqrt((1.00 x 10-3 W/m2)/(1/2(344m/s)(1.21kg/m^3)(12566.3^2)))
    A = 1.74 * 10 ^-7

    this seems wrong..
  11. Nov 10, 2006 #10

    Doc Al

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    Staff: Mentor

    I didn't check over your last calculations, but frequency and omega are related by: [itex]\omega = 2 \pi f[/itex]. Combine that with your equation from post #7.
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