0
$\begingroup$

We know an elastic medium is necessary for the propagation of any kind of wave like sound. But air is not necessarily an elastic medium at all. For the compression and rarefaction which is extremely necessary for the wave to flow a dragging force is highly required to bring the particle to equilibrium. But in case of air the normal motion of the particles is a completely random one without any kind of equilibrium position at all. What causes the rare fraction and compression and causes the wave to flow in air? And again the potential between two gas molecules doesn't even obey hooke's law at all. Then how can i treat the displacement of a gas particle at time t and space x like y=asin(wt-kx)?

On an attempt to solve this problem I thought to expand the potential between gas molecules,whatever let it be,in a Taylor series with parabolic approximation but that also causes problem because the equilibrium position is not set.

$\endgroup$
  • $\begingroup$ Nothing,i have no idea how to solve it.in all books all calculations are done treating air as an elastoc medium but i am wondering how it is possible.how can air be an elastic medium.any kind of help,mathematical or simulation will do. $\endgroup$ – user157588 May 29 '17 at 14:09
  • $\begingroup$ I've deleted the comment that you responded to because after re-reading your question, I see that my comment is not responsive. $\endgroup$ – garyp May 29 '17 at 14:15
  • 1
    $\begingroup$ You seem to be mixing the particle perspective with the bulk flow perspective. A single particle isn't a wave in this sense, it's a particle. $\endgroup$ – Kyle Kanos May 29 '17 at 15:31
1
$\begingroup$

Consider the Ideal Gas Law (which deals with the bulk properties of the air, not with specific molecular interactions, as eluded to in Kyle's comment).

If we use the "specific gas constant" version of the equation, which can examine density instead of a fixed volume, you can see that $$P = \rho \bar R T$$ where $P$ is pressure, $\rho$ is density, $\bar R$ is the specific gas constant, and $T$ is temperature.

You can see that density is proportional to pressure. Essentially, it means that as you try and compress the gas, it will build pressure.

As long as temperature is held constant, there will be a relationship between pressure and density that creates elastic behaviour. Specifically, when you add pressure, the density increases, when you remove pressure, it decreases.

This is how we describe elastic behaviour. Inelastic behaviour would be if it maintained it's density after pressure is lowered (and due to some temperature changes there actually would be some of this "elasticity" of motion that gets transferred to heat).

$\endgroup$
  • $\begingroup$ But the mathematical model y=asin(wt-kx) how does it work?the density may be dependent on pressure but there is no equilibrium position of particles which is necessary for a wave to flow $\endgroup$ – user157588 May 29 '17 at 16:54
  • $\begingroup$ @user157588 Equilibrium position is your average room pressure (~atmospheric). This pressure wave interacts with that, causing high pressure and low pressure zones that travel outward. $\endgroup$ – JMac May 29 '17 at 17:21
1
$\begingroup$

You don't need "elasticity", you need a restoring force proportional with displacement. And gas pressure provides such a force.

$\endgroup$
  • $\begingroup$ You basically said, it's not elasticity, it's (description of elastic behaviour). According to dictionary.com elasticity's definition in "physics" (definition 4) is basically exactly that, except instead of length volume or shape, it's density. Then the etymology dictionary gives an origin of "elastic" as "a scientific term to describe gases" in French. I went a little overboard on this comment because I was personally curious about the word (I wasn't actually sure what it meant). $\endgroup$ – JMac May 29 '17 at 20:20
  • $\begingroup$ @JMac yes. "You keep using that word. I do not think it means what you think it means." I thought OP needed the slightly different perspective $\endgroup$ – Floris May 29 '17 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.