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Acoustic power of loudspeakers and interference

  1. May 11, 2006 #1
    the problems is that:
    two loudspeakers, A and B, radiate sound uniformly in all directions in air at 20 degree C. The acoustic power output from A is 8x10^-4 W, and from B it is 6x10^-5 W. Both loudspeakers are vibrating in phase at a frequency of 172 Hz.

    a) Determine the difference in phase of the two signals at a point C along the line joining A and B, 3m from B and 4m from A.

    b) Determine the intensity and sound intensity level at C from speaker A if speaker B is turned off and the intensity and sound intensity level at point C from speaker B if speaker A is turned off.

    c) With both speakers on, what are the intensity and sound intensity level at C?

    In part a, i have tried to work out the y(x,t)=Acos(kx-wt) of both loudspeakers, then add them together, I have got the result 2A,
    but their phase difference is pi (3.14).

    In part b, i have no idea how to link the power and intensity together.

    thanks for attention.
     
  2. jcsd
  3. May 11, 2006 #2

    nrqed

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    There are several way to express this, but one possible way is the following: the phas difefrence at a given point (for two sources in phase) is simply the difference of distance to the two sources divided by the wavelength, all this multiplied by two pi, [itex] {2 \pi (r_2 -r_1) \over \lambda} [/itex] (where r_2 is the distance between the point and the farthest of the two sources). This makes sense, right? For example, if the point is equidistant to the two sources, the two waves are in phase at that point. If the point is at a distance [itex]\lambda[/itex] farher from one source than the other, there will be a difference of phase of 2 pi radians, etc.
    The intensity (in Watt per meter squared) is [itex] I = { P \over 4 \pi r^2} [/itex] where r is the distance between the point and the source and P is the power (in watts) of the source). As for the sound intensity level, it is [itex] \beta = 10 log {I \over 10^{-12} W/m^2}[/itex]. Notice that is is a log, not an ln. And it comes out in decibels.

    when the two sources ar on (assuming that they are incoherent), you must add the *intensities* of the two at the given point *before* calculating the sound level.

    Hope this helps.

    PAtrick
     
  4. May 12, 2006 #3
    thx for your helping,
    i will try my best to understand it
     
  5. May 12, 2006 #4
    i got a problem that how do you derive the equation of
    intensity of sound [itex] I = { P \over 4 \pi r^2} [/itex] ?
     
  6. May 12, 2006 #5
    and the answer in part c is wrong if i add the intensities at a particular point together.............
     
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