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The clarinet and the soprano sax are similar in size, but the clarinet is able to play much lower notes than the soprano sax. Why is this the case? I understand that when the clarinet is overblown the pitch goes up by a 12th, not an octave, which extends the range, but I don't understand the physics of how it can play so much lower than the soprano sax. It seems like it could be partly to do with the size of the mouthpiece (soprano sax has a much smaller mouthpiece) or possibly to do with the bore shape, but I'm not really sure whether any of these reasons are right.

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Both instruments (as well as all single- and double-reeded instruments and all brass instruments) behave as tubes closed at one end. The difference is that saxophone (and all other reeds) are conical, whereas the clarinet is cylindrical.

For a conical tube closed at one end, the fundamental wavelength is twice the length of the tube (Actually, it's a bit more complicated than that, but that will work for a basic understanding).

For a cylindrical tube closed at one end, the fundamental wavelength is four times the length of the tube (and the instrument only resonates odd harmonics). So for a basic approximation, a clarinet is going to be about an octave lower than a similarly sized conical instrument.

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    Just to complicate things :-), the transverse flute is essentially open near one end (not at the actual end of the head joint), and is more of a Helmholtz resonator than a tube resonator. But that wasn't part of the original question. Just putting this out in case someone wonders why the flute, with cylindrical bore, overblows to octave, not 12th. – Carl Witthoft Sep 17 at 15:46
  • @CarlWitthoft if the flute were a Helmholtz resonator then it wouldn't be able to cleanly overblow at all. The fact that it overblows the octave despite being cylindrical demonstrates that it indeed behaves as a tube resonator with both ends open. – leftaroundabout Sep 18 at 10:03
  • @leftaroundabout I'm not completely sure whether a Helmholtz device "counts" as an open-ended tube or not, but your point is well-taken. I have built crappy ocarinas which overblow, but that probably points more to lack of fipple quality than anything else. – Carl Witthoft Sep 18 at 11:55
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You pointed out the reason. The clarinet acts as a stopped pipe. An open pipe has a node (little air movement) at each end and an anti-node (lots of air movement) at the middle (or maybe I've got this backwards; check the physics of wind instruments pages.) Thus the clarinet's fundamental wavelength is twice the pipe length and the saxophone's is equal to its pipe length. Thus for the same length of pipe (the effective length of the instrument), the clarinet would sound an octave lower than the sax.

Next, (as mentioned in the question) the clarinet's natural overtones are the fundamental, third overtone, fifth, etc., while the sax's are the fundamental, second,third, etc. The clarinet starts and octave lower and its natural vibrations are the odd harmonics. A sax's natural vibrations are all the harmonics.

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  • You got it backwards: An open end does not allow for pressure to build up, so you always get nodes with movement but no pressure oscillation at an open pipe end. Both the sax and the clarinet have an open and a closed end (the mouth piece), and you get a pressure oscillation at the closed end of the clarinet. I.e. pressure oscillation at the mouth piece, and air movement at the other end. Which is the cause for the specific sound of the clarinet, because it means that its spectrum cannot contain any even harmonics. The sax behaves differently because of its conical shape. – cmaster - reinstate monica Sep 17 at 10:07
  • Is the difference because of the conical shape, or is the difference because of the reed design? A positive pressure wave pulse which hits an open pipe end will cause a negative pressure pulse to be sent back; a positive pulse hitting a close end will cause a positive pressure pulse to be sent back. A positive pressure pulse which pushes a reed "closed" will cause a negative pressure pulse back, while one that pushes it "open" will cause a positive pulse back. If a round trip results in a pulse of the same phase as the original, the pipe will support all harmonics with... – supercat Sep 17 at 18:56
  • ...a fundamental wavelength of twice the pipe length (one round trip). If it results in an anti-phase pulse, the anti-phase pulse will essentially cancel out any sound whose wavelength would be twice the pipe length, or any harmonics thereof, but it will support oscillation with a wavelength of four times the pipe length (two round trips), and odd harmonics thereof. – supercat Sep 17 at 18:59
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    @supercat It's the conical shape. Both instruments are blown with the same mouth piece. (I'm neither a sax nor clarinet player, so I won't rule out minor changes to the mouth piece shape, but the operating principle is exactly the same. And that's what matters). The conical shape makes the pipe behave like an open-open pipe, even though it's actually closed-open. – cmaster - reinstate monica Sep 18 at 10:52
  • @supercat The clarinet with the cylindrical bore and closed-open behavior produces the fundamental (four times its length), but cannot produce the even harmonics (second harmonic = twice the length) because that would require nodes of the same type at both ends of the pipe, which is not possible in the closed-open configuration. You find this behavior of missing even harmonics in other instruments as well, and it always tends to produce a very characteristic sounds that resembles the sound of a clarinet. – cmaster - reinstate monica Sep 18 at 10:59

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