# Relation of power and intensity of sound/point source wave with distance travelled

I have two conceptual doubts about 3D waves while self-studying,they are:

(1) For a 3D wave,we know that $$A \propto 1/r$$ Relations for intensity are as follows $$I \propto A^2 \ and \ I \propto 1/r^2$$ Shouldn't this imply that $I \propto 1/r^4$? Please tell me where am I messing up, as I can't figure out whats wrong here.

P.S.- I know the $A\propto 1/r$ comes from the intensity relations only and I have a feeling thats where im getting my logic wrong.

(2) By same relation $$A \propto 1/r$$ shouldn't the power ($p\propto A$) of a sound/point source wave decrease with distance?

• I believe you got a nice answer, but to help a bit more: if you replace the proportionality of amplitude to distance in the proportionality of intensity to amplitude you will end up with the I proportional to A squared relation, so your assumption that intensity is proportional to the fourth power of amplitude does not hold Jul 13, 2020 at 11:16

Remember that sound intensity is the amount of energy flowing through an area normal to the surface:

The surface area of a sphere is $A = 4 \pi r^2$. That's why intensity scales as the inverse square of the radius for point sources.

shouldn't the power (p∝A) of a sound/point source wave decrease with distance?

Be careful! This is one of the most common misconceptions for students starting out in acoustics. Power is a property of the source, and so it doesn't change with distance. Think about it this way: power is just the rate at which energy is being changed from one form to another.

If you're familiar with calculus, another way we can see how power doesn't change with distance is by looking at the following integral:

$$P = \oint \vec I \cdot \vec n \ dS$$

It says if we take the intensity over an entire closed surface (perpendicular to the surface) we'll get the power of the source. The energy of the source has simply been "stretched out" over a larger surface area with distance. Such an approach is convenient because as long as you don't have sinks of acoustic energy inside the surface you can quantify the total power of multiple sources inside of the surface - this is nice because most real-life sources (like an engine) are actually aggregates of myriad small sources of sound.

• thanks a lot! the calculus explanation and the elaborate diagram were very helpful! One thing to note is that I implied amplitude instead of area by 'A' however the fact that power remains constant helped clarify the other doubts too. Feb 19, 2018 at 15:45