Fundamental frequency of a rectangular-cavity whistle

  • #1
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Summary:

I'm making whistles of my own design on a 3D printer and they don't turn out with the frequencies I want
I designed a parametric CAD model of a whistle that can be 3D printed. Basically I designed the internal airspaces, put a skin around it, and printed it. Combine two of these in the same enclosing body, with slightly different frequencies, and you get a warbling sound similar to a pea whistle. This works as planned.

My whistle design uses a rectangular cavity. Typical dimensions would be 25mm long, 11mm wide, and 4mm deep, with a 4x11mm window at the top (air is blown onto the downstream edge of the window to drive the vibrations in the cavity).

I thought that because the cavity is open on one end (actually the end of one side) the fundamental frequency would be controlled by the overall length, but that isn't the case. Based on ##f=v/λ##, with the speed of sound ##v## as 343,000 mm/sec, 25mm of length would be a quarter-wavelength of 3430 Hz.

It's close... but the whistles always produce a tone a few steps lower than intended. If I change the depth of the cavity (say, change from 4mm to 3mm or 5mm) the pitch changes. Even worse, when I blow harder, the pitch changes -- one whistle goes up slightly and the other goes down slightly, removing the beating effect, so I suspect that something other than the cavity length becomes dominant.

So I suspect the fundamental frequency is based on volume, like the Helmholz resonance only my cavity isn't spherical and has no neck.

I found a formula for finding the fundamental frequency of a closed rectangular box which might be helpful except one end of my box is open.

Information I've found about resonant frequencies of open rectangular cavities are all academic papers that seem quite involved.

Is there a good simple approximation for finding the fundamental frequency of a rectangular whistle?
 
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Answers and Replies

  • #2
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Update: It occurs to me that maybe I can approximate this by using the formula for the resonant frequency of a tube open at one end:
$$ f=\frac{nv}{4(L+0.4d)}$$
where ##n=1## is the first resonance, ##v## is the speed of sound, ##L## is the cavity length, and ##d## is the effective diameter. I would calculate ##d## as the equivalent diameter of a circle of the same cross-sectional area; that is,
$$d=2 \sqrt{\frac{W L}{\pi}}$$
Solving for ##L## using a frequency of 3430 Hz, I get 22 mm, which is shorter than the 25 mm quarter-wavelength for that frequency. That might explain why my whistles have a lower pitch than I planned; my cavity is too long for the frequency I want.

Using this to calculate the frequency given that I already have a whistle that's 25 mm long, I get a frequency of 3063 Hz, which is still somewhat higher in pitch than the whistle, but maybe close enough for my purposes.

Another update: Looking further, I learned that the 0.4 term is known as an end correction because the pitch of a real tube is actually lower than the theoretical pitch. If I adjust this fudge factor to 0.5 the predicted frequency is quite close to what I'm observing, judging by comparing it by ear to this online tone generator.
 
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