I Sound pressure and inverse distance law

AI Thread Summary
The discussion centers on understanding the relationship between sound pressure and distance from a sound source, particularly how pressure follows an inverse distance law rather than an inverse square law. It clarifies that while intensity decreases with the inverse square of distance due to the increasing surface area of a spherical wavefront, sound pressure is a local measure based on the movement of air molecules and does not depend on the surface area of the wavefront. The area in the pressure equation refers to a unit area at a specific point, not the entire spherical surface. Consequently, the force exerted by sound waves changes with distance, while the area used in the pressure calculation remains constant. Overall, sound pressure is localized, contrasting with intensity, which is a more global measure related to the wavefront.
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Given spherical propagation of a sound wave. If pressure = force/area, and the increase in spherical surface area is proportional to r^2, why does pressure not decrease in an inverse square law, given that area is in the denominator of the pressure equation?
Hello,
I am confused as to how to think about sound pressure and the distance from a sound source. Let's assume that a sound wave is omnidirectional and propagating away from a source in a sphere. We have the following two equations:

Pressure = Force/Area
Intensity = Power/Area

Many texts I've read start with intensity and show that since the surface area of the sphere is increasing by 4πr^2, and the same quantity of power is distributed across the surface area, that the intensity is decreasing in a 1/r^2 fashion. This is the inverse square law. From there, the texts then explain that power is proportional to the square of Pressure and conclude that pressure then must decrease in a 1/r fashion. This makes sense mathematically. However, I'm confused as to how to think about pressure alone if I knew nothing about intensity and only had the pressure equation. In other words, instead of deriving the change in pressure from what I know about it's relationship to intensity, how would I derive the change in pressure if I knew nothing about intensity, and was just working with a concept of pressure? Here is where I get confused because the equation for pressure is:

Pressure = Force/Area

Given that Area is in the denominator of the pressure equation, I would assume an inverse square law to the change in pressure given the change in distance. However, given that pressure follows an inverse distance law, this must mean that Force increases with distance? Does force change over distance? Or is there a better way of thinking about this?

In short, given only the equation for pressure, how is it that pressure follows an inverse distance law and not an inverse square law? I know that area has r^2 in its equation (unless I'm incorrect when it comes to the area in the pressure equation), and area is in the denominator of the pressure equation. So, what is happening to the force and how should I think about this?

Thank you!
 
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Welcome to PF.

ngn said:
In short, given only the equation for pressure, how is it that pressure follows an inverse distance law and not an inverse square law?
The energy in the wavefront is attenuated by the inverse square law.

Work done = energy, is the product of a force by the distance moved.
The force is equivalent here to the pressure.
For a particular elastic material, the distance moved is proportional to the pressure.
Energy is therefore proportional to the square of the pressure or amplitude.

Pressure is reduced as the inverse of distance.
Power is the rate of flow of energy.
Power is reduced as the inverse square of distance.
 
Thank you for the response. I understand that the energy is attenuated by the inverse square law. I believe that is reflected in Intensity = power/area. I am wondering, however, about the pressure equation: Pressure = Force/Area. Does that equation not apply for pressure in a wave? If it does apply for a wave, then doesn't the surface area increase in proportion to radius^2? If so, why does the pressure not drop in proportion to radius^2, given that Area is in the denominator of the pressure equation?
 
ngn said:
If so, why does the pressure not drop in proportion to radius^2, given that Area is in the denominator of the pressure equation?
Because pressure is measured as a force per unit area, and that one unit area does not change with radius.
 
I think you have pinpointed my source of confusion! My confusion is that when I look at the equation for pressure (Force/Area), I assumed the Area was the surface area of the spherical wavefront. I realize that this is incorrect now. With sound, the pressure is based on the local movement of air molecules, not a measure that regards all of the molecules contained within a sphere.

So, I have a some questions.

1. When I see the equation for sound pressure (Pressure = Force/Area). What is the Area referring to? Is it the area of a single molecule of air? Is it a single point in space? Is pressure a local measure and not one that considers the entire spherical wavefront?

2. Is it correct to assume that this Area for pressure does not change at all based on distance from the source? It is thus the force alone that changes based on distance?

3. When it comes to calculating intensity (Intensity = Power/Area), then that Area is indeed the surface area of the spherical wavefront, correct? Since the surface area of the sphere increases proportional to radius^2, this explains the inverse square law for intensity.

4. When you mention Work = Force x Distance. What does the distance refer to? Is it the distance from the source? Is it the local distance that each particle is displaced? Or this local distance summed across all particles? Does it only pertain to the wave front?

5. Am I right in assuming that if we were trying to calculate the pressure of air in a spherical container, then we would calculate the force over the surface Area of the sphere? In this case, the pressure would decrease proportionally with the container's radius^2? However, with sound, that is not the right conceptual framework regarding the Area?

Overall, I have a hunch that pressure is local and intensity is global in relation to the wave, but I need some help to solidify that idea.

Thank you again for the help!
 
Last edited:
ngn said:
1. When I see the equation for sound pressure (Pressure = Force/Area). What is the Area referring to?
Using SI units; pascal = newton/metre2.
In customary or imperial units; psi = pounds/inch2.
ngn said:
2. Is it correct to assume that this Area for pressure does not change at all based on distance from the source?
Yes.
ngn said:
3. When it comes to calculating intensity (Intensity = Power/Area), then that Area is indeed the surface area of the spherical wavefront, correct?
Yes.
ngn said:
4. When you mention Work = Force x Distance. What does the distance refer to?
A force acts on an object. If the object moves in the direction of the force, then work is done on the object by the force.
Energy = force * displacement.
ngn said:
5. Am I right in assuming that if we were trying to calculate the pressure of air in a spherical container, then we would calculate the force over the surface Area of the sphere?
Pressure is always force per unit area. The pressure is independent of the surface area of the container. Pressure will change if the volume of the container changes.
 
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