The way I see it,if there's a set up A which has 5 particles and another set up B which also has 5 particles,and assuming everything else is same in these set up,like distance between particles etc,if I increase the amplitude of particles in A set up,then that particle would go and push the 2nd particle,but,since it's amplitude has increased,it and the 2nd particle together will go and hit the 3rd particle(assuming the particle have enough amplitude to achieve this),similarly,now the 2nd and 3rd particle should do the same,so,since one particle is pushing 2 particles,this,should increase the speed compared to the set up B,so why doesn't this happen?
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2 Answers
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Increasing the kinetic energy of molecules in a gas means the temperature has to grow.
It is true that the speed of sound grows with temperature, see the measured change:
So the average kinetic energy rising increases the speed of sound.
See also the related question and answer here .
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$\begingroup$ So u mean to say when I increase the amplitude while keeping the frequency constant,the particle moves faster and it's speed increases as it covers more distance in same time,as it's displacement is more but since the frequency is constant,it will cover that more distance in the same time,so,this means it's kinetic energy has increased and it has now become a hot air or whatever medium and indeed,the speed in hot air or medium is more,am I right? $\endgroup$ Commented Mar 2, 2022 at 7:14
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$\begingroup$ @AakashGarain No, If you can understand the physics formulas, the increase is in the mean kinetic energy" , the root mean square of velocity. Particles do not move over the whole volume the way you describe but back and forth in the gas in average locations by hitting each other in the gas, raising the kinetic energy when heated to higher temperature ( to get the amplitude increase you are trying to describe) $\endgroup$– anna vCommented Mar 2, 2022 at 7:46
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1$\begingroup$ What I mean to say was,I know that the particle doesn't move over to the whole side,the particle moves back and forth,it vibrates,and I meant to say that it vibrates faster,right?I mean,the frequency will be the same,but,since the amplitude has increased,it is covering more distance during 1 vibration,and so,it's speed has increased during vibration,right? $\endgroup$ Commented Mar 2, 2022 at 8:26
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$\begingroup$ It is statistical mechanics applied to gases, one does not talk of speeds but of average kinetic energy which depends on the average speed, In a gas molecules are not tied up to other molecules as in liquid and solids to speak of vibrations around a point. $\endgroup$– anna vCommented Mar 2, 2022 at 8:47
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$\begingroup$ in solids is what vibrates locally but you are talking of free particles hitting each other that is why I gave the gas example .Solids phys.libretexts.org/Bookshelves/University_Physics/… $\endgroup$– anna vCommented Mar 2, 2022 at 8:55
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You are mixing up two speeds, the speed at which "information" is transmitted from one molecule to the next (wave speed) and the speed of oscillation of a "molecule" about a "mean" position.
I must now clarify something which is very important.
The information speed is related to the "thermal" speed of the molecules.
As all molecules are travelling at different speeds one needs to consider some sort of average speed of the molecules which in turn dictates the time of travel between molecular collisions.
"The speed of oscillation of a molecule about a mean position" without qualification is a nonsensical statement.
What I mean by that is that one must consider the motion of the centre of mass of a small volume of gas about a mean position when the sound wave is not present.
Thus it is an aggregate of the motion of a lot of gas molecules in a small volume with the molecules inside that volume constantly changing.
So now to do a comparison.
At room temperature, the average thermal speed of air molecules is about $500\,\rm m/s$.
The intensity of a sound wave is given by the equation $I = \frac 12 \omega^2 A^2 \rho c = \frac 12 v_{\rm max}^2\rho c$ where $\omega$ is the frequency, $A$ is the displacement amplitude of a molecule about its mean position, $\rho$ is the density, $c$ the speed of the wave and $v_{\rm max}$ the maximum speed of a molecules when executing its oscillatory motion.
For a normal conversation the sound intensity is approximately $10^{-6}\,\rm W/m^2$ and using $330\,\rm m/s$ for the speed of sound and $1.2 \,\rm kg/m^3$ for the density of air gives $v_{\rm max} \approx 10^{-5}\,\rm m/s$, a value which is considerably smaller than the thermal speed of the molecules.
The oscillatory motion of the molecules due to the passage of a sound wave can be thought as a small, but significant, perturbation of the thermal speed of a molecule.
Thus the maximum speed of oscillation of a molecule about a mean position is very much less than the speed at which the information, the motion of one molecule about a mean position being communicated to its neighbour, is being transmitted.
For solids I think that it is much easier as to what is happening in that the molecules only vibrate about a fixed position in the lattice and the information is transmitted via the bonds between the molecules.
