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I have a general understanding of how everything works. I understand standing waves, and the difference between open- and closed-ended tubes.

I understand that for closed-ended tubes, the tube must be in the order of 1/4, 3/4, 5/4 (etc...) of a wavelength in order for the sound to occur. This creates the antinode at the open end.

Likewise, I understand that for open-ended tubes, the tube must in the order of 1/2, 1, 3/2 (etc...) of a wavelength in order for the sound to occur. This creates the antinodes at the open ends.

All my textbooks (and everything I find online) simply says that when the antinodes line up like this, a sound is heard. I get it, but is this the only way a sound is made in these air columns? Why is it that whenever I blow sideways into a bottle (regardless of where the liquid is, or how long the bottle is) I can make a sound? Textbooks seem to simply indicate that a sound is only heard when the antinodes line up with the open ends.

My guess to my answer is: Sure, sounds can be heard at all lengths of tube. However, when antinodes line up at the open-ends, the sound is resonated. Is that correct?

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  • $\begingroup$ When you blow sideways over the mouth of a bottle, you're exciting a Helmholtz resonance. It's not a longitudinal vibration of an air column. $\endgroup$ – user4552 Oct 26 '19 at 18:56
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Like the rest of the contributors have already stated here, you can excite the air column (the air in the tube, bottle, etc.) with a broadband source, or a "monochromatic" source.

In case this is a continuous source (i.e. does persist in time) we are talking about what is usually found as forced vibrations (like nasu stated in their comment). In this case, the frequency of the oscillation will match that of the source. If this frequency happens to coincide with one of the resonant frequencies of the system (tube, bottle, etc.) the amplitude will be maximised (whether the radiated energy is maximised or not depends on other parameters too - such as the surrounding medium, the shape of the vibrating system, the impedances of the radiating material as well as the surrounding medium, etc.). If you are talking about a tube with an open end, the "shape" of the vibration will exhibit a maximum at the open end.

In the case, the driving frequency does not coincide with a resonant frequency then the system will still vibrate, but with lower amplitude and even at the open end the amplitude will be non-zero. Considering that, pretty much always some energy will be radiated (due to non-zero or non-infinite terminating impedance at the open end), but in non-resonant cases, the amplitude will vary and will increase when closing to a resonant frequency.

Now, when you excite the system and consequently let it vibrate freely, like nasu mentions in their answer, it will pick the resonant frequencies (and not just one, but most probably a superposition of as many as possible) and the rest will decay very fast. Most probably the resonant frequencies will exhibit different decay rates too, based on the material and the internal damping (or even the radiated energy at each frequency, since this constitutes another mechanism with which the system loses energy). This is why you get very characteristic peaks from guitars. While on the contrary from a flute you also get some "noisy" parts in the spectrum from the forced vibration due to the broadband source (the airflow from the lips). But in the latter case (flute) those frequencies that are not supported by the system (non-resonant) exhibit very small amplitude.

Finally, you have to keep in mind that most of the time the resonant frequencies will be somewhat off from the calculated ones (in the ideal case) due to the terminating impedance. Unless this is included in the calculations (formulation/modelling of the system), most often the resonant frequencies will be somewhat lower than the calculated ones due to the "added mass" of the air at the open end.

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Yes, your understanding is correct - the textbooks refer to resonance. You could put a speaker next to a column of air and make sound of any frequency - but some frequencies will interfere constructively after reflection from the other end of the column.

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Regarding the sound made by blowing sideways in a bottle, you can make it regardless of what is the level of the water but the pitch of the sound will depend on this level. The air column will vibrate such that one of the "resonant" modes is selected. It is similar to a guitar string. Sure, you can pick any string and force it to vibrate to any frequency you want, with an attached vibrator for example. But once you leave it to vibrate freely it will soon "pick-up" a resonant mode. The waves with other wavelengths are damped much faster than the ones that can make standing waves. A similar behavior applies to tubes. See for example the Helmholtz resonator for more detail.

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  • $\begingroup$ So are you saying - if I blow into a tube (or bottle) with a frequency that doesn't make it resonate, it will make a sound, but will dampen very quickly (as soon as I stop blowing, I guess). However, if I change the frequency (or the length of the tube by adding water, for example), so that the standing waves produce an antinode at the open end, it will resonate and make a song that will last longer? $\endgroup$ – Jumpman Jul 2 '17 at 2:09
  • $\begingroup$ When you blow in the bottle you don't "blow with a frequency". Do yo think you can adjust your breath to frequencies of hundreds of Hz? It is the same with exciting the guitar strings. You don't pick them "with a frequency" but just remove them from equilibrium and let them go. When you blow you change the pressure in the bottle and it gets back to equilibrium pressure oscillating with some frequency. There is a difference between free oscillations and forced oscillations. $\endgroup$ – nasu Jul 2 '17 at 5:12
  • $\begingroup$ Right. I guess what I meant was, "However, if I change the frequency (by instead of blowing, using a tuning fork of different frequency) or change the length of the tube (by adding water), so that the standing waves produce an antinode at the open end, it will resonate and make a sound that will last longer?" $\endgroup$ – Jumpman Jul 2 '17 at 7:24
  • $\begingroup$ Yes, using a tunning fork is different from just blowing. $\endgroup$ – nasu Jul 2 '17 at 14:17

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