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Looking at wind instruments, there is an apparent relation between the size of the instrument and the pitches it can produce; The larger it is, the lower pitches it produces.

The question I have is, is it, without changing the type of material used, possible to change the range of a wind instrument by just changing its shape? (i.e. without changing its volume)

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  • $\begingroup$ usually a longer/wider instrument can create lower frequency standing waves and, thus, a lower pitch $\endgroup$ – Jim Aug 10 '15 at 15:50
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Yes. The simplest change one could make to change the range is to modify the end condition. A pipe based instrument with an air reed has an open end at the air reed. The other end could be either open or closed. If that end is closed, the range goes down an octave, and the overtones change (so the tone, or timbre, changes, too).

Also, changing from a cylindrical shape to a conical or flared shape will change the range. Here is some info for a tube from a table in Rossing's The Science of Sound.

$$\matrix{\text{Mode} & \text{Cylindrical} & \text{Conical} & \text{Flared} \\ 1 & 136.4~ \text{Hz} & 255.6~ \text{Hz} & 281~\text{Hz} \\ 2 & 409.2 & 512 & 520 \\ 3 & 682.0 & 771.0 & 767 \\ 4 & 954.9 & 1032.0 & 1021}$$

The reason for the change is that the pressure nodes in the conical and flared tubes are non-linearly distributed.

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As others have already pointed out, changing the length, and changing the boundary conditions (open end/closed end are simplest...) will change the frequency of the fundamental frequency. For an open-open pipe (like a flute where it is open at the end that you blow over and open at the first uncovered hole) the wavelength of the n^th harmonic is just

$\lambda_n = \frac{2L}{n}$

where L is the length of the tube.

But you can also change the shape of the tube. The above formula assumes that the tube is a simple cylinder. If it was a cone, or some other shape the formula is more complicated. In real life many wind instruments (bassoons, oboes, saxophones,...) have conical bores instead of cylindrical bores.

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Of course it is: the pitch depends on the length of the tube, not the volume. Roughly speaking, a tube 40 meters long and 1 cm diameter will have a fundamental frequency one quarter that of a tube 10 meters long and 4cm diameter.

I may be off by half a wavelength here (open-ended tube resonant frequency).

The bore of wind instruments both in diameter and variation of diameter over its length, is designed to favor certain overtone amplitudes.

EDIT: to clarify, if you freeze the tube length and diameter, simply bending it into a weird shape (looking at you, French Horn!) can change the overtones but not the fundamental.

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