Why does sound pressure vary with 1/r?

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SUMMARY

Sound pressure decreases with distance according to the inverse relationship of 1/r, while sound intensity diminishes with the square of the distance (1/r^2). This phenomenon occurs because sound propagates spherically, causing the intensity to spread over a larger area as distance increases. The discussion highlights the analogy of gas behavior to a compressed spring, emphasizing that pressure waves can be understood through the principles of force and energy in compressible materials.

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  • Basic knowledge of physics principles, particularly force and energy
  • Conceptual grasp of compressible fluids and their behavior
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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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