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Magnitude of acoustical image

  1. Oct 26, 2005 #1
    In basic acoustic thoery a sound source reflects across a boundry in a fasion similar to light. The magnitude of the reflected wave can be determined by adjusting it for the reflection loss and the distance traversed.
    I am looking at a scientific paper from the 70's. In it they use the following equation to caculate the magnitude of a wave emanating from an image x meters away.
    y = \frac {1} {x}
    My intuition would tell me that the magnitude of a wave emanating from an image as a function of distance would be described by the following:
    y = \frac {1} {4 \pi x^2}
    Does anyone know if the former equation is really correct? Why is the inverse square law not used?
    Last edited: Oct 26, 2005
  2. jcsd
  3. Oct 26, 2005 #2


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    the magnitude of any radiation intensity, that is something measured in watts/m^2 must be inverse square (assuming no transmission loss) to satisfy the conservation of energy axiom.

    however for sound, the intensity of radiation is the product of the RMS pressure, which is 1/r for a spherically expanding wavefront, and the RMS of the in-phase component particle velocity, which is also 1/r. it turns out that the quadrature component of the particle velocity (that is the component that is 90o out of phase with the pressure wave) is 1/r^2, but that component does not contribute to power transmission. so if both pressure and in-phase particle velocity are 1/r, then their product is 1/r^2 or inverse square.

    there is another thread here called Inverse Square Relationship that you might want to check out. if not, do check out




    the first link talks a little about acoustic waves and 1/r vs. 1/r^2
  4. Jul 3, 2007 #3
    So is the pressure of the sound wave [tex]P = \frac {1} {4 \pi r}[/tex]?? Or is it [tex]P = \frac {1}{r}[/tex]?? The references don't seem to make it terribly clear.
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