This document discusses how wind instruments produce sound through air vibrations forming standing waves inside cylindrical tubes. It explains that most wind instruments can be approximated as open-open or open-closed tubes. The document compares how the flute and clarinet differ as open-open and closed-open tubes respectively, affecting their possible note ranges. It also explores how instrument length impacts the lowest playable frequency and how an octave relationship is formed between instruments like the flute and piccolo. Sample problems demonstrate how leaks or breaks could limit an instrument's range by changing the effective tube length.

1. PITCH AND RANGE
IN WIND INSTRUMENTS
By Yong Jia Bu
Physics 101 203

2. • music produced on wind instruments come from the vibrations of air
through the instrument
• we can approximate most wind instruments as cylindrical tubes
through which air is blown
area of high pressure area of low pressure
• this may look very much like normal sound waves in air, but because
they are confined to the cylindrical tube (or air column), vibrations of
air inside the instrument actually become standing waves
• To understand these standing waves, we need to first look at how air
at the ends of the air column interact with atmosphere

3. Flute: open-open tube
Clarinet:
closed-open tube
It is probably easy to understand that the clarinet is closed at one end (mouthpiece
completely covered by player’s lips when playing), but it may not seem obvious that
the flute is an open-open tube. When playing, a flautist does not cover the hole in
the mouthpiece entirely, but blows air across the mouthpiece. Air can still come in
contact with atmospheric air this way.
• most wind instruments are of either the open-open tube or the open-
closed tube type
• open ends are parts of the tube where air inside the tube makes
contact with the outside, atmospheric air

4. • It is relatively easy for a player to change the air pressure inside the
wind instrument by blowing air into it
• it is NOT easy to change atmospheric pressure outside the instrument
• as vibrations in the instrument reaches the inside air and atmospheric
air interface, it is almost as though the vibrations “hit a wall” and then
sends sound waves in all directions
• the waves that are reflected back into the instrument after bouncing
off the atmospheric “wall” now travel through the instrument in the
opposite direction
To audience

5. • standing waves are formed when two waves of the same amplitude,
wavelength, and frequency travel in opposite directions and interfere
with each other
Fixed end Fixed end
node
Anti-node Anti-node
• standing waves only form if they can vibrate along a length of
medium that is some integer multiple of ½ of the wavelength;
otherwise, when the wave is reflected, the amplitude doesn’t quite
add up to something that stays constant!
• the pitches (frequencies) that wind instruments can produce depend
on the length of the instrument

6. Fixed end Fixed end
• The fundamental frequency of a note produces a standing wave like
the one below:
• the fundamental frequency has no nodes between the fixed ends, and
it isn’t possible to fit any longer wavelengths into the medium without
disrupting the standing wave
• Think of how instruments with lower ranges are
larger, longer, so that they may play notes with
longer wavelengths
Bassoon
Range: A1–E5
Flute
Range: C4–C7

7. Open
(hits atmosphere)
Open (hits
atmosphere)
• Open-open tube, e.g. flute: both ends open to the atmosphere, a
wave is sent out every time the vibrations reach either end
Open (hits
atmosphere)
Closed
(no vibrations
sent out)
• Closed-open tube, e.g. clarinet: one end is a closed, dead end with
zero air movement (air pressure is always at a maximum at this
end because air has nowhere to expand)
Longest possible wavelength λ = 2*length of tube
Fundamental frequency f = v/2*length of tube
Longest possible wavelength λ = 4*length of tube
Fundamental frequency f = v/4*length of tube
This shows why the clarinet can play in a lower range than the flute
although the two are about the same length.

8. • A note that is an octave above another
note has exactly double the frequency of
the other note
C5: f = 523.23 Hz
C4: f = 261.63 Hz
• The piccolo is designed to be an instrument that plays
everything at one octave higher than a regular flute. If we
measure the distance from the piccolo’s mouthpiece to its
bell, we can see that this distance is about ½ of the
distance on flute (ignoring end corrections). The piccolo’s
fundamental frequency is therefore an octave higher.
Piccolo
Flute

9. 1. Suppose you accidentally dropped your flute during practice and
one of the keys has gotten twisted so it no longer closes completely.
What would this mean in terms of the range of notes you can/can no
longer play (assume the flute is a simple open-open air column and
ignore modern designs on the instrument that change its
properties)?
2. Because your flute is broken you are practicing your clarinet
instead. But there seems to be a leak on your clarinet as well. You
identify that the leaky key is 2/3 down the length of your clarinet
(measured from the mouthpiece). Assuming that your clarinet is a
simple closed-open air column of length L, calculate the difference
in frequencies between the original lowest note that you could play
and the lowest note that you are able to play while there is a leak
(the answer is in terms of L).

10. 1. If the flute is a simple open-open air column, then its length is
effectively only as long as its last closed key. Theoretically, if there
is a key in the middle that refuses to close, air can contact the
atmosphere at this leaky point, making it the new “fixed end” of the
air column and rendering all keys past it useless. And so notes with
wavelengths greater than two times the length of the mouthpiece to
the leaky key are no longer playable. The closer the leaky key is to
the mouthpiece, the fewer notes will be left playable.
2. A closed-open tube has the fundamental frequency f = vair/4L, so the
normal fundamental frequency should be about 86L Hz. The new
fundamental frequency is f = vair/(4*(2/3)L), which is 129L Hz. The
difference between the original frequency and the new frequency is
43L Hz.

11. Wolfe, Joe. Poetics. “Open vs. Closed Pipes (Flutes vs. Clarinets).” The University
of New South Wales School of Physics. The University of New South
Wales, n.d. Web. 7 Mar. 2015. <http://newt.phys.unsw.edu.au/jw/
flutes.v.clarinets.html>.