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Saxophones, oboes and generally generally most woodwind instruments I can think of behave like open pipes. That seems logical; when every key is closed, they actually are open pipes and, if I remember correctly, a pierced pipe behaves like a (slightly longer) open pipe.

The clarinet, however, behaves like a closed pipe. This has two major consequence I can think of: the partials of a clarinet songs consists of odd harmonics (at least in first approximation); a clarinet register key makes the instrument play a twelfth (thrice the fundamental) instead of an octave (twice the fundamental).

What is the physical explanation behind this phenomenon?

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3 Answers 3

The clarinet has a cylindrical bore, which makes it behave as a closed tube, odd harmonics.

See http://en.wikipedia.org/wiki/Bore_(wind_instruments)

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I guessed that it’s related to it’s bore, but why would a cylindrical clarinet still behaves like an stopped cylinder pipe instead of an open one? Isn’t it an open pipe? If it behaves as closed on the mouthpiece end, why don’t other reed instrument do the same? –  Édouard Jan 27 '14 at 22:08
The flute also has a cylindrical bore and behaves like an open pipe - this does not help much (and the wikipedia article also offers no additional explanation). –  guidot Jan 28 '14 at 12:47

Saxophones and oboes are conical, and behave like closed conical pipes. They are closed at the reed, just like the clarinet.

Flutes are cylindrical, and behave like open cylindrical pipes. The sound is made by blowing across the opening at the head joint, and it is not closed like in other woodwinds.

Clarinets are cylindrical like the flute, but closed at the reed, so they behave like closed cylindrical pipes.

Each category has its own set of unique characteristics. You can find a detailed explanation here: http://www.phys.unsw.edu.au/music/ . It goes more in depth than most sources, and requires that you are comfortable with the math involved in a typical university physics course.

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When I asked this question to a math professor, his answer was: just because the differential equations modelling the air column resemble those of a closed tube. All other explanations offered here do not help much (at least in the presented summaries), since the bassoon, also with a conical bore and a stopping reed at the end overblows into the octave, and therefore also behaves like an open tube. So I would summarize to "there is no easy explanation, dive into advanced math".

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I'm not sure what you're saying here; the bassoon "behaves like an open tube" because it has a conical bore. The basic rule is that open tubes and conical-bore instruments behave similarly and stopped cylindrical-bore instruments behave differently from them; understanding why that's true requires a lot of math, but applying it doesn't. –  Micah Jan 28 '14 at 19:14
What I‘m trying to say is: 1) cylindrical or conical does not decide alone - see flute (where everything but the head is cylindrical). 2) The construction of the mouth-piece - however closed it looks to the eye also does not cut it - saxophone and clarinet are quite similar in that respect. So simple recipes like taking that bore and that mouth-piece will give that tube behavior still seems ambitious to me, even if one can learn it for existing instruments easily. –  guidot Jan 28 '14 at 20:45
What's relevant is the combination of the mouth-piece and the bore. A closed mouthpiece and cylindrical bore lead to "clarinet-like" behavior (overblowing at the twelfth, small even harmonics, lower fundamental); any other combination leads to "non-clarinet-like" behavior (overblowing at the octave, large even harmonics, higher fundamental). So a flute is non-clarinet-like because it has an open mouthpiece (despite being cylindrical) while a saxophone is non-clarinet-like because it has a conical bore (despite having a closed mouthpiece). –  Micah Jan 28 '14 at 21:50

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