Why does sound pressure vary with 1/r?

AI Thread Summary
Sound pressure decreases with distance according to the inverse relationship of 1/r due to the nature of pressure waves propagating through a medium. While sound intensity diminishes with the square of the distance (1/r^2) because it spreads over a larger area, sound pressure reflects the amplitude of the wave rather than the energy distribution. The analogy of a gas acting like a compressible spring helps illustrate this concept, as the pressure correlates with the force exerted by the gas particles. Understanding this relationship clarifies why sound pressure does not follow the same squared distance decay as intensity. Overall, the behavior of sound pressure is rooted in the physics of wave propagation and the properties of compressible media.
Bitruder
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I didn't follow the template because this is more of a conceptual question that I can't get a clear answer for.

I understand that sound intensity varies with 1/r^2 because the total intensity at a point in the wave is constant and if you have spherical propagation then the area of the surface of that sphere increases with r^2.

I can also calculate pressure from the formulas of intensity that include pressure.

But can anybody give an intuitive explanation for why pressure drops off with distance and not distance squared? Since sound is a traveling pressure wave, I suppose we can pretend that we are riding a pressure peak out in space along a ray.

Thanks
 
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The force of a compressed spring is F = kx and the energy in a compressed spring is kx^2/2

Can you convince yourself that a gas acts like a compressible spring?
 
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