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Amplitude decrease with geometrical spreading 
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#1
Aug114, 10:55 PM

P: 12

Hey,
I'm looking at amplitude decrease of a seismic pulse as a result of geometrical spreading. Starting with I = E / (4 * pi * r^{2}) where E = original energy from source, we know that energy falls off as 1/r^{2}, 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} = 4r^{2} (which is 4x r^{2}). 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
Aug314, 10:18 PM

Sci Advisor
HW Helper
P: 6,684

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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