# Sprinting and Sound

Congratulations to all the Olympic athletes who have competed thus far, especially Michael Phelps, Nastia Liukin, and Shawn Johnson. Gymnastic events are all great to watch, and I don’t think you could find a more colorful analyst than Bela Karolyi. The balance beam is probably the most classic and thrilling gymnastic event – the Summer Olympics equivalent of figure skating – but the men’s high bar and women’s uneven bars are just absolutely jaw-dropping. It looks like magic.

Now the track meets take center stage. My younger brother ran track in high school, and he tells me the track event he has the most respect for is the 800 meter race. As he puts it, “It’s long enough that you can’t run the entire race at full speed, but it’s short enough that you have to run the entire race at full speed.”

Sounds brutal. I’ve never done anything more than casual jogging, and I expect my 800 meter time would be more like 4 minutes than the Olympic standard of somewhere in the 1:40s. I’m in reasonable shape, but I’m definitely no sprinter.

Anyway, enough Olympics. While we’re talking about speed though, we have a good excuse to do a toy calculation and look at some numbers.

Temperature is a measure of the average kinetic energy of the molecules of a substance. The actual kinetic energies of each molecule will vary about that average. We do know that the way to convert from temperature to energy is to multiply the temperature by Boltzmann constant k. Room temperature might be taken to be 25 C, so convert that to Kelvin and multiply by k and you’ll get E = 0.0257 eV. We know how much a nitrogen molecule weighs, so we can use the kinetic energy relationship to find the velocity:

Plugging in the mass of the nitrogen molecule, I get v = 445 m/s. Now this is a ludicrously loose calculation. I’ve completely ignored the Maxwell-Boltzmann distribution, issues involving degrees of freedom, and the fact that air is not entirely nitrogen. But you might notice nonetheless that this is not too far from the speed of sound, which at that temperature is about 346 m/s. You might expect the speed of sound to be about the same as the speed of the air molecules, since it’s the motion of the air molecules that carry the sound waves.

It further turns out that if we replace the 2 in the above equation with 7/5 (the adiabatic index) to correct for the degrees of freedom of the diatomic nitrogen, we get a speed of 373 m/s which is actually a pretty darn good approximation considering the roughness of the calculation.

It’s one of those things that make you feel good when you’re figuring out thermodynamics – yes, all these weird concepts and calculations do in fact give you numbers which aren’t too far removed from what we actually see in everyday life.

1. #1 Brian
August 22, 2008

Bait and switch!

2. #2 Robert E. Harris
August 22, 2008

Mean translational K. E. is 3kT/2. Root mean square (rms) speed is square root of 3kT/m, mean (arithmetic average) is square root of 8kT/((pi)m). Sorry, I don’t have the fonts and tools for nicer presentation. At 298K, values for N2 are 515 m/s (rms) and 475 m/s. The adiabatic index has nothing to do with it. Internal degrees of freedom are not connected with the center of mass motion, so far as these speeds are concerned. The adiabatic index comes into the speed of sound because sound travels as a series of (nearly) reversible adiabatic compressions and rarefactions.

Nice to see a post on this subject. Kinetic theory of gases is an interesting subject in itself. If you can find Martin Knudsen’s little book on it, read it. It’s full of Knudsen’s neat experimental results. REH

3. #3 camello
August 22, 2008

silly question: what does the average speed of a molecule have to do with the speed of propagation of sound waves? they seem completely independent to me.

4. #4 CCPhysicist
August 23, 2008

Very disappointing. Given a chance to do a nice back of the envelope calculation on the importance of the speed of sound in sprinting events, you dropped the baton as badly as that alleged “team” of US sprinters. (Had they ever run together before at full speed?)

The speed of sound is a big deal in track! How long would it take sound to get from the gun to the guy in an outer lane for the 200 or 400 m, not to mention the 4×400, if they didn’t use speakers to level the playing field across the staggered start? (Did I mix that metaphor enough?) Too long.

There is a similar issue from the geometry side when it comes to being DQ’d for running on the line. Some commentary I heard implied it was a technicality, as if it were only about interfering with the other runner — but 0.01 s at 10 m/s is only 10 cm. Make the track 20 cm shorter by running a smaller radius, and that 0.02 s can move you up a couple places.

5. #5 Matt Springer
August 23, 2008

For #3, generally a disturbance can’t propagate faster than the things being disturbed. If you have three molecules 1 2 and 3, number 1 is deflected by the incoming wave and goes to hit 2, which then travels in turn to hit 3. Each molecule has to cover the distance between it and the next molecule, which can only happen at the speed the molecules are traveling. So since sound is the disturbance going from one molecule to the next, sound will be going pretty much at the speed of the molecular motion. There are complicating factors, but the two speeds are pretty close.

#2, the equipartition theorem guarantees that the internal degrees of freedom do have just as much energy as each of the three translational degrees of freedom. I think you were discussing along those lines, but I want to make sure that gets across to the readers.

Good points, CC, but I can’t cover everything. Gotta leave something for the other blogs. ðŸ˜‰

6. #6 Robert E. Harris
August 25, 2008

Internal degrees of freedom obey equipartition in classical kinetic theory. N2 is of course a QM vibrator, so the separation of the vibrational levels in comparison with kT is important. At 25C the value of kT is about 200 cm-1, vibrational frequency for N2 is about 2300 or 2400 cm-1, so population of first vibrational excited state is very small, about 0.001 % of population of ground vibrational state; second is about zero, and vibration contributes just about nothing to heat capacity at room T. Rotational levels are close enough spaced that N2 has the full kT of energy for the two rotational modes. Hence the adiabatic exponent of 7/5.

Sorry about the spectroscopists’ cm-1 here, just ones I can remember. Multiply cm-1 by hc with c in cm/s to get regular units. (I don’t off hand know a conversion to eV.)

Point about molecules needing to collide to carry sound in gas is a good one. Doesn’t work in condensed phases. But as my old boss (George Jura) used to say, “The gas phase is very complicated. Only the liquid and solid phases are more complicated.” Or words to that effect.

I’m enjoying your posts.

You can learn a lot from bad teachers if you figure out how to present the material yourself, which you ought to do to get the most out of any class.

REH

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