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Speed of an air molecule when transmitting sound of frequency f 
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#19
Jul1009, 11:46 AM

P: 219

Yes, but there is the extra variable x in the last equation you mentioned.
I need the proof that A does not depend on f, or if it does, how. Anybody knows any more advanced physics forums? 


#20
Jul1009, 11:55 AM

P: 512

The velocity is omega*sqrrt.(A^2x^2) After all that explanation tell me what tells you there is a link between A and f? The average velocity is the average of the maximum velocity which is at the centre, A*omega and the minimum velocity 0 at the Amp.,A*omega/2 which is at x=sqrrt.(3)/2A and not x=1/2A as it is not uniform but oscillatory motion. You got the relation between velocity and frequency How do I get you a relation between A and f if it does not at all depend on f? 


#21
Jul1009, 01:13 PM

P: 416




#22
Jul1009, 01:28 PM

P: 219

I told you what is the conclusion I am interested of. It is the title of this thread. I want to conclude (or to reject) that the higher the frequency, the higher the average speed of an air molecule that its collisions cause this sound, whereas the speed of that sound remains the same no matter how high the frequency is. You are saying that this is obvious. Man, it is not obvious, otherwise why so many people say that the speed of sound is equal to the speed of the molecule?
The average speed of the molecule is 4Af. Therefore if A and f are not dependent at all, then this conclusion is correct. But I fear whether A gets smaller as f increases, so that the speed of the air molecule does not increase as the frequency increases. I don't see any clue that suggests that A and f depend on each other, otherwise I would have an indication of how to solve the problem. But I also can not be certain that they do not depend on each other. If I do not see it prooven clearly through the relative equations, I cannot be sure. In that simulation I mentioned, A and f are shown as independent, but that simulation has the mistake that the speed of the firts wavefront (the speed of sound), increases as f increases. I think I read somewhere that this indeed happens in the case of the sound in solids, but in the air its speed does not increase as f increases. One more clue: If the speed of the molecule increases as the f increases, then how come and the temperature of the air does not increase? One answer is that sound is not causedtransmited by all the molecules of the air which this sound traverses. Is this true though? Something is wrong in all this. When A or/and f is zero, the molecule already has a speed. So its average speed is not 4Af? But IT IS 4Af. 


#23
Jul1009, 03:28 PM

HW Helper
P: 7,107

The speed of air molecules is much faster than the speed of sound.
For the specific example of dry air at 20°C, the speed of sound in air is 343 m/s, while the rms speed of air molecules is 502 m/s using a mean mass of air molecules of 29 amu : http://hyperphysics.phyastr.gsu.edu...d/souspe3.html more links: http://www.ems.psu.edu/~bannon/moledyn.html http://en.wikipedia.org/wiki/Acoustic_velocity Getting back to the original question, what I don't know is if the average kinetic energy of the air (temperature) is affected by sound waves. Do sound waves just organize the otherwise random movements of air molecules without any change in average kinetic energy (temperature)? I'm also not sure how shock (super sonic) waves, which have the low pressure part of the wave clipped at zero pressure (negative pressure doesn't exists in the real world), affect the air. 


#24
Jul1209, 04:57 PM

P: 219

502m/sec is their resultant velocity (when not transmiting sound). Thus its component velocity at each of the three axises should be (502m/sec)/3^(1/2)=289.8m/sec. Correct?
Their speed at the direction of sound is only one of the 3 component velocities, at only one of the 3 axises xyz. Correct? (This is also weird if it implies that the speed of sound is greater than the component velocity of the molecule at the direction of sound. But I guess it does not imply that, but that after sound starts, the 502 velocity increases, so the 502 refers to before the sound starts?) And that component velocity should be 4Af. THIS IS IMPOSSIBLE if A and f are independent. It means that when it does not transmit any sound, f=0, thus its speed is zero at one of the three axises. Correct? ANY PHYSICISTS IN THIS FORUM? THERE MUST BE SOME! 


#25
Jul1209, 09:46 PM

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P: 7,107




#26
Jul1309, 12:12 AM

P: 1

If the question is how fast the molecules of air are moving, here's a web page that may be of interest. http://www.newton.dep.anl.gov/askasc.../chem03448.htm But, let me quote a passage of interest here: "There's a really neat mathematical equation based on a theorem called the "equipartition theorem" which states that the energy of a gas system (equal to 1/2*mv^2) is equal to the temperature of the gas (equal to 3/2*kT). If we rewrite this equation to solve for velocity we get: sqrt(3*T*k/m) = v where T is the temperature in Kelvin, k is the Boltzman constant = 1.3805*10^ 23 J/K and m is the mass of the gas particle. If we assume that the average mass of air (since it is a mixture of different gases) is 28.9 g/mol (or each gas particle is around 4.799*10^26), and room temperature is 27C or 300K, we find that the average velocity of a single air particle is around 500 m/s or 1100 miles per hour!" The reaon I find this particularly interesting as it relates to the speed of sound is by thinking of what the average speed might be in a linear direction. If an air disturbance is propelled by the collisions of the molecules and the molecules are moving at a nominal 1100 mph, some of the time the sound will be propagted at that speed. But, related to that selected direction, some will be at right angles to that direction and will propagate along that axis at zero mph. Other rates will depend on other angles and should average out to about the speed of propagation at 45 degrees. That puts the average at about 770 mph along any given axis. To me that seems just a little too close to the nominal Mach 1, under standard conditions, of 761 mph to be a simple coincidence. 


#27
Jul1309, 02:46 AM

P: 512

study of sound transmission requires "no wind", it disrupts wave 


#28
Jul1309, 03:01 AM

P: 512




#29
Jul1309, 03:12 AM

P: 512




#30
Jul1309, 03:36 AM

P: 512

This is going far from the simple oscillatory equations to the kinetic theory. You just have tto superimpose the average initial velocity and the proposed velocity of particle oscillations to get the new velocity.



#31
Jul1309, 06:42 AM

P: 219

According to their logic the average speed is 4Af and the the max speed is 2πAf.
The average speed is wrong, thus the max speed most probably wrong for the same reasons. Because when f tends to zero, the speed does not tend to zero. Also, if we consider A to be the actual displacement of the molecule, then the factor A is correct. But we consider that when there is no sound, A is zero, thus the factor A is also wrong. I guess these equations would be correct if the molecule was still when there is no sound. The molecule is making a difinite number of oscillations (which is not definded by f) within a definite distance (which is not 2A) in each sec, when there is no sound. Therefore, someone must find the correct equation. If I am wrong show me why. My mistake could be that we can consider the speed of the molecule as zero when there is no sound, but whyhow? Vin, don't say again that the average speed is zero "in all directions", because according to this logic, the speed in the above equations should also be zero. 


#32
Jul1509, 03:48 PM

P: 101

"A Textbook of Sound" 3rd Rv. A.B Wood cpyrt 1955. G.Bell and Sons Ltd. Pg 248  249 "velocity of small amplitude waves" excerpts [tex]\sqrt{\overline{u^2}}=482 metres/sec[/tex] = root mean square velocity ; [tex]\bar{u}=447metres/sec[/tex] = mean velocity of molecules The Kinetic Theory of Gases shows very simply that the pressure in a gas is given by [tex]P=\frac{1}{3}Nm\overline{u^2}=\frac{1}{3}\rho\overline{u^2}[/tex] where [tex]\\N[/tex] is the number of molecules per c.c., [tex]\\m[/tex] the mass of a molecule, and [tex]\rho[/tex] the density of the gas. This relation expresses the molecular velocity in terms of the pressure and density of the gas, thus [tex]\overline{u^2}=\frac{3P}{\rho}[/tex] Since the energy of the motion of the molecules in a given (i.e. xaxis) we may write speed of sound = [tex]\sqrt{\overline{u_{x}^2}}=\sqrt{\frac{P}{\rho}}[/tex] Which is Newton's "Isothermal Velocity" of wavepropogation in the gas. If we employ the mean molecular velocity [tex]\bar{u}[/tex] instead of [tex]\sqrt{\overline{u^2}}[/tex] we find [tex]\sqrt{\overline{u^2}}=\bar{u}\sqrt{\frac{3\pi}{8}}[/tex]  Pg 275276 "Velocity of Sounds of High Frequency  in Gases" excerpt H.O. Kneser, in a number of theoretical and experimental papers dealing with anomalous absorption and dispersion of sound, has derived the following expression for the velocity in terms of frequency, [tex]\\w=2\pi\\N[/tex], and molecular constants: [tex]\\v=\sqrt{\frac{P}{\rho}}\left(1+R\frac{c_{v}+w^2\beta^2c_{va}}{c_{v}^2 +w^2\beta^2c_{va}^2}\right)[/tex] where [tex]\\c_{v}[/tex] is the molecular heat at constant volume, [tex]\\c_{va}[/tex] is the specific heat of the translation degrees of freedom, [tex]\\R[/tex] is the universal gas constant, and [tex]\beta[/tex] is the mean life of the energyquantum, that is the time involved in the quantum transformation  translationalintramoleculartranslational energy. at low frequencies the velocity becomes [tex]\sqrt{\frac{P}{\rho}}\left(1+\frac{R}{c_v}\right)[/tex] whereas at high frequencies it becomes [tex]\sqrt{\frac{P}{\rho}}\left(1+\frac{R}{c_va}\right)[/tex]  Anyways it appears to me that you should be able to substitute around and solve for molecular velocity 


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