4

I understand the basic physics of a Swanee whistle: shorter tube -> higher pitch.

But when I take the head joint off my flute, play a note and poke my finger up the tube, this behaves in exactly the opposite way. The further in my finger is, the lower the note. At first glance, it seems to be shorter tube -> lower pitch. I want to know why this is.

For the record, I'm not changing my embouchure and the note pitch changes way more than I can bend a note with my lips or by turning the mouthpiece - I'm definitely not using any trickery to do this.

I've only just discovered the physics of musical instruments page from this question, and I think it might be something to do with closed- vs open- pipes. I know a flute (and its head joint, presumably) behaves like an open pipe. So maybe my question should really be "Is a Swanee whistle a closed pipe?". I think the swanee whistles that I've seen have a mouthpiece similar to a recorder or penny whistle - you blow in the top and the air is split slightly lower down - I'm not sure of the technical term for that part of the instrument.

  • 1
    I think fipple is the term. And yes - swanee whistles ARE like that. It's an interesting question and I hope you'll get a good technical answer soon. I'm just worrying you'll end up in hospital with your finger stuck up a flute. – Old Brixtonian Sep 17 at 20:50
4

The crucial thing is that poking your finger just a little way into the tube does not close it off – it's still an open-pipe situation, unlike an extended swanee whistle. Only as you push it in further, the remaining opening gets smaller and smaller, thereby changing the acoustic behaviour from “open pipe” to “closed pipe”, and the closed pipe has a much lower resonance (full octave lower, ideally speaking).

Note that if you close it off with your flat hand instead, which does form a seal, then the pitch goes immediately down low – lower than when you push your finger in.

| improve this answer | |
  • 1
    When you completely seal the end, the fundamental resonance is no longer the vibration in the tube, but the Helmholz resonance at the embouchure hole. Vibration within the tube is a function of tube length, but the Helmholz resonance is a function of the hole size, the tube thickness, and the volume of the closed space. It's going to be much lower, just like blowing across the mouth of a bottle. Any audible frequency from the tube would essentially be an overtone. – Tom Serb Sep 17 at 21:43
  • @TomSerb Helmholtz resonance is also vibration in the tube. But what I'd understand as Helmholtz is the situation where the opening is much smaller (orders of magnitude) than the cross-section of the body, which is not really the case for flute or recorder. But yeah, it is significantly smaller, so the result is somewhere between half-open tube and Helmholtz resonator. – leftaroundabout Sep 17 at 21:54
  • Yep. There's no step between Helmholz resonance and open tube resonance. With experimentation, it's easy to find a position for your finger up the tube with a small opening that has exactly the same pitch as closing off the tube at the very end. – Scott Wallace Sep 18 at 14:44
  • The opening doesn't have to be orders of magnitude smaller - any opening will do. The sound hole on a guitar is a Helmholtz resonator. – Tom Serb Sep 26 at 14:20
  • @TomSerb it's not really Helmholtz either, but indeed the approximation works better in a guitar because there the volume is quite big from both a width and height, not just from a single length dimension. – leftaroundabout Sep 26 at 18:46

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.