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I would like to clear up some confusion about the mechanics of air particles that are propagating a sound wave.

I understand that there is no net movement of air molecules when a sound wave passes through air. Instead, the particles oscillate and the wave is propagated through various elastic collisions between air molecules which cause the compression to keep moving forward.

What I don’t understand is how the air molecules move back to approximately their original position after colliding with the other particles to keep the wave moving further. Doesn’t this seemingly violate the laws of conservation of momentum? (This can’t be the case since sound exists) If a particle hits its neighbor, and that neighbor molecule now has the momentum from the wave to keep moving forward, how can the original particle have the momentum to move backward and to its original place.

I would also like to clarify that I understand gas molecules have their own random motion in addition to the wave motion and was wondering if this had something to do with the aforementioned phenomena.

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  • $\begingroup$ I think I got ur problem : u wanna know that since molecules are of same mass and collision is perfectly elastic. So the first particles going to collide should stop at the moment it collides . Right ?? $\endgroup$
    – Ankit
    Jun 17 at 10:06
  • $\begingroup$ Sound in air (or liquid) travels as a P wave, and has net momentum along its propogation direction. Ultrasound emission can be 'weighed' by a delicate balance. The air motion in sound is not entirely symmetric oscillation. $\endgroup$
    – Whit3rd
    Jun 18 at 6:54
  • $\begingroup$ Perhaps they do NOT. Oscillating air molecules make a wave. $\endgroup$
    – user196418
    Jun 18 at 11:57
  • $\begingroup$ Yes @Ankit, that is one of my concerns. Whit3rd, does that mean that simple harmonic motion shouldn’t be used to describe the motion of particles in a sound wave? $\endgroup$
    – Srihari P
    Jun 19 at 5:29
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Sound waves are referred to as pressure waves, and if you understand this, it should answer your question.

When sound waves propagate, they form alternate regions of high pressure, called compressions, and low pressure, called rarefactions in the air. The air molecules move toward and away from these regions as the wave propagates. It is not that momentum is not conserved (momentum is always conserved), but more about the motion of particles moving into lower pressure regions, and those that move away from higher pressure regions, as the wave travels through the air.

Also, the motion of individual gas molecules in the air is pretty much random, but as explained above, as the sound waves travel through the air, there is a collective motion of gas molecules in the volume surrounding the vibrating sound wave.

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Without the random movement of molecules, it is true that a vibrating membrane for example would create a depleted region around. It is like a fight in the middle of a crowd. Suddenly there is a clear region around the event and a wave of pressure that extends for some distance.

But because of that random movements, any depleted area is filled almost immediately. The same reasoning is valid along the propagation. The net displacement of molecules in the wave front is like a vibrating membrane pushing the next layer, and the random movements fill the gaps behind.

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Your question is a question on elastic collisions, disguised as a question on wave mechanics:

What I don’t understand is how the air molecules move back to approximately their original position after colliding with the other particles to keep the wave moving further. Doesn’t this seemingly violate the laws of conservation of momentum?

The fact that, after impact, one particle moves forward and one backward does not imply a violation of momentum conservation. To understand why you have to remember that momentum is defined as a vector, not as a scalar. Suppose the first particle, with velocity $\vec{v} _i$, impacts another still particle ($\vec{u}_i=0$); surely conservation of momentum applies:

$$m_1\vec{v}_i+m_2\vec{u}_i=m_1\vec{v}_f+m_2\vec{u}_f$$

where of course the subscripts diriminate between the quantity before impact ($i$) and after impact ($f$). In our case we get:

$$\vec{v}_i=\vec{v}_f+\vec{u}_f$$ since $u_i=0$ and the particles have the same mass. The final velocity of our first particle ($\vec{v}_f$) is then: $$\vec{v}_f=\vec{v}_i-\vec{u}_f$$ and now we can see that the conservation of angular momentum allows $v_f$ to be positive or negative, depending on who is bigger between $v_i$ and $u_f$.
Furthermore: if the impact was perfectly elastic we should have also considered conservation of kinetic energy and this extra condition would have imposed opposite particle velocities after impact! For example: think about billiard balls impacting each others: those are nearly perfect elastic impacts, and in fact after impact you never see billiard balls moving in the same direction.

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  • $\begingroup$ Thank you! This makes a lot of sense. One point of confusion I still have is that because energy and momentum both have to be concerned, does this mean that the particles when moving backwards are slower than they were prior to making a collision or that the wave dissipates really quickly? $\endgroup$
    – Srihari P
    Jun 19 at 5:25
  • $\begingroup$ You can run the numbers and find out for yourself. It's just a group of 2 equations. $\endgroup$
    – Noumeno
    Jun 19 at 7:28

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