7
$\begingroup$

This question is not about the theory but about getting a moderately accurate (ideally 1%) estimate of the speed of sound in the current conditions. The range of conditions that I am interested in are those in which humans can live without special support.

I found this in Wikipedia:

$$c_{air} = 331.3 \sqrt{1 + \frac{\theta}{273.15}}$$

where $\theta$ is the temperature in Celsius.

I am slightly surprised that the pressure does not feature. Is this because the pressure is determined by the temperature alone? My knowledge of meteorology is poor but I would have expected that it is possible to have the same temperature at different pressures and that this would affect the speed of sound.

How about humidity?

I want this for a light hearted experiment. Some of my musical friends are able to judge small fractions of a semitone in musical intervals. I want to see how accurately we can judge the speed of a vehicle by the Doppler shift in a sound that it is producing.

For example, if the speed of sound is currently $343.2ms^{-1}$ (sea level at $20C$) and the apparent note drops by a (well tempered) minor third then the vehicle was travelling at $106.8kmh^{-1}$.

I have not yet allowed for wind speed, that will be a later refinement.

$\endgroup$
  • $\begingroup$ Some useful? data here. kayelaby.npl.co.uk/general_physics/2_4/2_4_1.html $\endgroup$ – Farcher Jul 31 '17 at 11:37
  • 1
    $\begingroup$ @Farcher Thanks, that is very useful. It tells me that the effect of humidity is well below my 1% requirement. I don't think even my friends who are best able to judge relative pitch could be that accurate. The information on more exotic gases is interesting but more that I need. It is curious that deuterium is listed but not regular hydrogen. $\endgroup$ – badjohn Jul 31 '17 at 11:49
  • $\begingroup$ It is interesting that Hydrogen is omitted. I have always found Kaye and Laby a very useful source for such information. $\endgroup$ – Farcher Jul 31 '17 at 11:52
  • $\begingroup$ I think you might be surprised at how good human short-term pitch memory is. I can play a note on a piano, walk over to a guitar & play what should be the same note and generally know if they are in tune by well under 1Hz (based on, say 220Hz (open A on a guitar)). This works for only a few seconds as I forget the note I played. I don't have perfect pitch (or anything like it) and I don't think I am exceptional in any way -- in fact I would assume that proper musicians can do significantly better -- I've just done it a lot over many years. $\endgroup$ – tfb Jul 31 '17 at 17:13
  • $\begingroup$ @tfb Well, finding out how good they are is one of my objectives. Actually, perfect pitch or pitch memory is not required. Even if we could identify the frequency of the note (when the vehicle was approaching) it would not tell us the speed unless we knew the true note. However (if my calculations are correct), we only need the interval between the note when the vehicle is approaching and receding. Any good musician should be able to do that. I can to within a semitone and maybe add a comment such as "a slightly flat perfect 4th". That does not give me 1km/h accuracy. $\endgroup$ – badjohn Jul 31 '17 at 17:24
11
$\begingroup$

If you take your equation:

$$ c_{air} = 331.3 \sqrt{1 + \frac{\theta}{273.15}} $$

We can rearrange it to get:

$$ c_{air} = 331.3 \sqrt{\frac{273.15 + \theta}{273.15}} = 20.05\sqrt{T}$$

Where now $T$ is the temperature in Kelvin. The formula for the speed of sound in an ideal gas is:

$$ v = \sqrt{\gamma\frac{P}{\rho}} \tag{1} $$

where $P$ is the pressure, $\rho$ is the density, and $\gamma$ is the adiabatic index. For an ideal gas we know:

$$ P = \frac{nRT}{V} \tag{2} $$

where $n$ is the number of moles of the gas, and the density is:

$$ \rho = \frac{nM}{V} $$

where $M$ is the molar mass in kilograms. The point of all this is that we can substitute for $n/V$ in equation (2) to get:

$$ \frac{P}{\rho} = \frac{RT}{M} $$

and substitute in equation (1) to get:

$$ v = \sqrt{\frac{\gamma RT}{M}} $$

For air $\gamma = 1.4$ and $M=0.0288$ kg/mol, and substituting these values into our equation for $v$ gives:

$$ v = 20.10 \sqrt{T} $$

which is the same as your equation give or take some rounding errors. That's how your equation was arrived at. As you suspected the pressure is involved, but the pressure and density cancel each other out in such a way that the speed depends only on temperature.

Humidity will have an effect because it changes the density of the air and it changes the average molecular weight. You can calculate this, but I would simply Google for empirical equations giving the speed as a function of both temperature and humidity.

$\endgroup$
  • $\begingroup$ Thanks. I could, and should have, managed that myself. My Googling has not worked yet but I will try harder. $\endgroup$ – badjohn Jul 31 '17 at 11:24
  • $\begingroup$ Together with the data from the link that @farcher provided, I seem to have the information that I require. Quite convenient, I need only consider the temperature which is fairly easy to measure. $\endgroup$ – badjohn Jul 31 '17 at 11:52

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.