Sound wave energy dissipation

[Dogoth]

I Was in a discussion with a fellow recently about sound (well general physics but the rules still apply). I said that if you had an omni-directional sound source (or as close to it as practical) the energy dissipation would be plotted on a inverse log π scale. He said no it would be log sq.

Firstly this is something I've believed for a long time and to my poor pee brain makes sense. In one dimension, if you throw a pebble into a still pond, the waves spread out in a circle dissipating their energy into more and more resistance as the circle expands. This resistance grows at a log π rate so wouldn't the inverse of that be how you would plot the energy loss. I understand that if the energy was directional (I.E. contained and could only dissipate linearly), then the law of log sq would of course apply. Neither of these scenarios would really exist in the real world so this is a theoretical question.

How do you plot the dissipation of energy of an omni-directional source radiating waves into in a free field?

I tried google but didn't get a satisfactory answer.

Thanks in advance for your reply

Cheers

[Mikeo]

Inverse square law

Pic: if I have your question right.

http://hyperphysics.phy-astr.gsu.edu/hbase/Acoustic/invsqs.html

 

I think I'm the only one here on HC that has a degree and Physics.

 

I just happen to see you question here.

 

 

[Dogoth]

Thanks Mikeo.

Yes the discussion was with an old friend who has a masters EE degree so rather than refute him, I'm in search of why.

The diagram was helpful but seems wrong. Looking at that sphere, there is a curve (I'm speaking of in the physical world, not mathematical world) to both the original and the expanded spheres that isn't taken into account. As a sphere grows in size it's surface area growth would be plotted on a log π scale correct? As sound waves propagate in a spherical fashion, isn't the energy diffused into a ever growing spherical medium (for our discussion air). Even the diagram doesn't draw them as squares. It appears that each graduation is effecting a larger surface area than would be plotted by using sq as the exponent. It still looks like π to me. What am I missing in my thought process? Here's the formula I found for deriving the surface area of a sphere from it's radius. Shouldn't this dictate the energy dissipation for any wave radiating from the center point? https://www.allmathtricks.com/sphere-surface-area-volume-hemisphere/

All of the books say I'm wrong. I'd just like to understand why.

[Dogoth]

OK I figured it out (I love these thought experiments).

To derive the surface area of a sphere the formula is: ( r )  = 4 x ( π r² ) = 4 π r²   To derive the volume of a sphere  the formula is: 4/3π r³

Both of which would seem to be useful in understanding power dissipation of the aforementioned waveform HOWEVER......... If you start with a single point source and begin radiating equally in all directions, the pattern has already become spherical no matter where you test it (1 meter or 1 micron distance from source makes no difference). That being the case (here's the pun 🙄 ) π is already baked into the recipe. You are measuring a single point of an expanding ball at a growing r (radius) from the source so, based on that straight line, of course log² would apply.

I know this only relates to audio in the measure of SPL and is barely relevant in real world practice so thank you for letting me post such an odd topic here.

My brain is satiated but if you have anything to add feel free.

Thanks again for allowing my public brain storming.

Cheers

[daddymack]

dammit...now my brain hurts...I got a C- in differential calculus...

[FreestyleIntruder]

The sound intensity (in watts per m^2) is equal to the sound power W of the source divided by the area S of the wavefront.

 If we're talking about a point source (omnidirectional) in free-field conditions then it's emitting sound waves in a sphere around it, therefore at distance r from the source I = W/(4 . pi . r^2)

This is often referred to as geometric spreading.

Linear or planar sources do things slightly differently.

The sound intensity *level* (in dB) is given by 10log10(I/I0) where I0 is the reference sound intensity of 1x10^-12 W/m^2