MadSci Network: Physics
Query:

Re: Why do the oboe and clarinet resonate differentlyt?

Area: Physics
Posted By: Dave Dixon, Assistant Research Physicist,University of California
Date: Wed Apr 9 20:21:38 1997
Message ID: 859765927.Ph


Unfortunately, there doesn't seem to be an easy answer to this question.
So, let me try and explain why the closed cylindrical pipe and the
conical pipe are different, and if you're feeling mathematically ambitious,
you can check the references below for the rigorous answer as to why the
conical pipe resonates with the same frequencies as the open cylindrical
pipe.

Charles Taylor, in his book "Exploring Music" (see below), makes a nice
analogy for the vibrating air column in a cylindrical pipe.  Imagine a
set of toy train cars in a line on a track, each attached to its neighbors
with a spring.  The case where the cars on either end are unattached
corresponds to an open pipe.  If you push on one end, you set up a
compression wave in the chain, which presses the springs together as
it travels.  When the wave hits the car on the other end, it simply
moves out as far as the spring will stretch, since it has nothing to push
against.  This then pulls on its previous neighbor, which sets up
an expansion wave moving in the opposite direction, where now the
springs are stretched as the wave travels.  Then, when the expansion
wave hits the car on the starting end, the opposite happens, and it
is reflected back again as a compression wave.  Since there's friction
between the cars and the track, if we want to keep this up, we need
to keep pushing on the system.  For the lowest frequency, we thus would
push again when the wave returned to the first car.  The situation in
an open cylindrical pipe is the same, except the compression and expansion
waves are in air.  Also, we see why the wavelength of the fundamental
is twice the length, since the wave must make one trip up the pipe and
another back before we push again.

The analogy for a closed pipe would be to attach one end of the
string of train cars to an immovable wall, again using a string.
Now, when we start a compression wave on end, it also reflects
as a compression wave from the other end, since the last car can
push against the wall.  This compression wave returns to the free
end, where it reflects as an expansion wave, reflects from the
fixed end as an expansion wave, and finally reflects back from the
free end, once again as a compression wave.  So the wave must travel
up and back TWICE before it returns to its original state.  If we
wanted to sustain the vibration, we thus would have to wait twice
as long as in the case with two free ends before we pushed again,
and so the wavelength of the fundamental is four times the pipe length.
Also, we see why the closed pipe has only odd harmonics.  Suppose
we tried to generate a frequency that was twice the fundamental,
by pushing after the wave had made one round trip.  After one trip
down the pipe and back, the wave is an expansion wave, put we're
pushing in as a compression wave, and the two would cancel each other
out.

Now, the reason why the train analogy works so well for a cylindrical
pipe is because of the cylindrical shape.  Since it has the same width
all the way down, the air in the pipe basically has no place to go but
forward and backward.  For a conical pipe, though, the air also gets
to expand outwards, and that complicates things significantly.  We
might get an approximate idea of the effect if, for our set of train
cars above, we didn't use springs which all had the same stiffness,
but rather decreased the stiffness as we went from one end to the other.
The waves in this case definitely will not be the same as in the ones
above, because different cars will move different amounts, etc., depending
on their position in the chain.  Likewise, in a conical bore instrument,
because the air can move inwards and outwards (from the center) as well
as back and forth, the wave pattern in the air column is NOT the nice,
simple sinusoid for a cylindrical bore, as you often see in textbooks,
but rather is distorted.  It works out that the particular way this
distortion occurs for a conical bore allows it to resonate with the
same frequencies as an open cylindrical pipe.

I wish I had a better answer, but was unable to find one that didn't
require solving the wave equation for a conical tube, and that's complicated.

References:

Charles Taylor, "Exploring Music: The Science and Technology of Tones and
Tunes", Institute of Physics Publishing, 1992 ISBN 0-7503-0213-5 (A very nice
book; contains the train analogy I gave above)

John M. Eargle, "Music, Sound, and Technology", Van Nostrand Reinhold, 1995
ISBN 0-442-02034-1 (A bit more sophisticated; takes a different angle 
on the cylindrical vs. conical problem)

Cornelis J. Nederveen, "Acoustical Aspects of Woodwind Instruments".
No publication information.  I think the publisher is Frits Knuf, 1969.
(This is fairly technical, and gives the derivation of the resonances
of a conical bore via solution of the wave equation).


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