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Amplitude decrease with geometrical spreading

  1. Aug 1, 2014 #1
    Hey,

    I'm looking at amplitude decrease of a seismic pulse as a result of geometrical spreading.

    Starting with I = E / (4 * pi * r2) where E = original energy from source, we know that energy falls off as 1/r2, thus amplitude falls off as 1/r.

    From wikipedia: "The energy or intensity decreases (divided by 4) as the distance r is doubled;"

    This makes sense to me, as when r is doubled we have the energy divided by (2r)2 = 4r2 (which is 4x r2).

    From this same principle, I would expect that the amplitude is divided by 2 when the distance is doubled as we have 1/2r instead of 1/r.

    However from a Louisiana State University website:

    "Geomteric spreading makes the amplitude of a signal falls off in proportion to the distance traveled by the ray. So that if the path of flight is doubled the amplitude will decrease by a factor of: square root of 2."

    I can't see how they got their factor of √2 instead of 2.

    Is one a mistake or am I missing something?

    Cheers
     
  2. jcsd
  3. Aug 3, 2014 #2

    Andrew Mason

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    Think of a compression/sound wave spreading out from a source equally in all directions as concentric spherical shells of energy. The energy in a given shell is constant. So the energy density is inversely proportional to the area of that sphere: i.e. energy per unit area varies as 1/r^2 where r is the radius of the shell.

    The question you are asking has to do with the relationship between amplitude and energy density of a wave front. Think of the vibration of a spring: the energy contained in the spring is proportional to the square of the maximum amplitude. PE = kx^2/2 .

    Since the energy contained in the wave front is proportional to the square of the amplitude and the energy density is inversely proportional to r^2, how would amplitude vary with r?

    AM
     
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