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The definition of sound (acoustic) intensity is given by

$$ I = {p^2 \over {\rho c}} \;\;\;\; \text{or} \;\;\;\; I = {p^2 \over {2\rho c}}$$

I've seen both definitions in different textbooks and am not sure which equation is more accurate. But in either case, the relation of intensity and density of air is:

$$ I \propto {1 \over \rho} $$

This contradicts my intuition.

I would hear no sound in a vacuum because there is no molecules in the medium to scatter the vibration from the source. If I start adding some molecules in the medium (low density), the source will vibrate these molecules in the medium, acoustic energy can be transferred and sound may be heard if there are just enough molecules hitting by eardrum (and within the audible frequency range). If I add a whole lot more molecules in the medium (high density), many more molecules will vibrate and hit my eardrum, and I would therefore hear a louder sound (or detect a higher acoustic intensity).

So why the equation suggests the opposite relation? Could it be that the instantaneous pressure is also a function of density, or $ p(t, \rho) $, in a way that pressure is proportional to density of higher order to offset the drop of $ 1 \over \rho $ have on intensity?

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A candid way to visualise that $density / amplitude$ relationship is the following image:

  • Imagine the medium as a 1 or 2 or 3 dimensional lattice of marbles linked to their nearest neighbors by springs (a N-dimensional array of harmonic oscillators, if you wish).

  • The density (of the simplistic medium) is related to the mass of the marbles, consequently, for a given (fixed) amount of energy transferred to the medium, the amplitude (of the oscillations) will diminish for different (increasing) values of the mass, to keep kinetic energy constant.

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  • $\begingroup$ That's a great way to visualize for someone with a shaky physics background like myself. But then again, if the same energy is transferred to more "marbles", then each marble will share less energy then that in a less dense system, and less energy should be transferred to my eardrum - higher the density, lower the intensity, and my ear hear a softer sound (which is not what we experience). So the contradiction is still there for me unfortunately. $\endgroup$
    – KMC
    Jun 29 at 2:23
  • $\begingroup$ Yes, but on the other hand, the higher the capacity of a mechanical system to store energy (as kinetic energy), the harder it is to set in motion. $\endgroup$ Jun 29 at 9:11
  • $\begingroup$ If it is harder to set in motion, then I should hear a softer sound in a dense medium (which actually agrees to the equation $ I \propto \rho $). But in reality, I should surely hear a louder sound if a medium is dense ($ 1 \propto 1/\rho $). $\endgroup$
    – KMC
    Jun 29 at 9:41
  • $\begingroup$ If you picture somebody playing the same tune in exactly the same way on two guitars, one with light gage strings and then on a acoustic bass (with the same body, etc.) with very heavy strings, the sound coming out of the bass will have a lesser amplitude, because the strings are harder to set in motion. $\endgroup$ Jun 29 at 9:49
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Your intuition is in the right direction. Nevertheless, you are only considering the amplitude and the intensity is the power per unit area. So, you need to consider that when a material is more dense you need a higher increment in pressure to get the same increment in particle speed.

If we write it mathematically, we get the following relationship between the pressure and particle velocity

$$v = \frac{p}{\rho c}\, $$

where $\rho c$ is the characteristic acoustic impedance of the medium. Thus,

$$I = \frac{1}{2}\operatorname{Re}\{p v^*\}\, ,$$

or

$$I = \frac{1}{2}\frac{|p|^2}{\rho c}\, .$$

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  • $\begingroup$ So what that essentially implies is for a given point in space (or a distance away from an acoustic source), I'll have to increase the power of the source in order for that point to observe/detect the same pressure amplitude if the density of the medium increases? $\endgroup$
    – KMC
    Jun 28 at 17:43
  • $\begingroup$ How do you go from the first equation to the second equation, and why you take the complex conjugate of particle velocity? $\endgroup$
    – KMC
    Jun 28 at 17:59
  • $\begingroup$ That's the definition of acoustic intensity, check this. Intensity is, indeed, a vector quantity but usually you are just interested in the magnitude. The complex conjugate is used when writing a complex representation of the waves. $\endgroup$
    – nicoguaro
    Jun 28 at 18:34

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