[Angelo Campanella]

"Dick Pierce" wrote in message
...

Another way to look at is that a closed-end forces the
particle velocity to be zero, and the pressure to be
at a max, while an open end allows the particle velocity
to be at a max while the pressure is zero. That means that
a closed end will provide a pressure node and a velocity
anti-node while the open end provides a velocity node and
a pressure anti-node. In a standing wave, adjacent pairs
of nodes or antinodes are found a half wavelength apart, while
alternating nodes and antinodes are found 1/4 wavelength
part.

OK

So we can look at the combinations of the two: a pipe with
two open ends or two closed ends provides a matching set of
nodes, and can support standing waves of all multiples of
half wavelengths: 1/2 wave, 1 wave, 1 1/2 wave, etc.

Thus, assuming the speed of sound is about 1200 ft/sec, an
open or fully stopped pipe 6" long can support resonances
at 500 Hz, 1000 Hz, 1500 Hz, etc.

whoops...

1200 ft/second sound speed has 1200Hz sound as a wavelength of one foot.

600 Hz sound will have a wavelength of two feet, so a half-wave will be one
foot or 12".
1200 Hz sound , will have a half-wave (closed ends resonator) of one-half
foot or 6".
Harmonics of hat closed tube being 2400 Hz, 3600, 4800.

On the other hand, a pipe which is open at one end and stopped
at the other supports a node at one end and an anti-node at
the other supports resonance that are odd multiples of 1/4
wavelengths.

OK

The same 6" pipe, stopped at one end, open at
the other, support resonances at 250 Hz, 750 Hz, 1250 Hz, ......

A one-half foot length of tube closed at one end will have a quarter wave
resonance for a two foot wave, or 600 Hz.
Harmonics are at odd # of quarters, or 600, 1800, 3000, ......

And yet another way to look at the behavior at the end of an
open pipe is that it forms an acoustic inertance (acoustic
mass) whose magnitude is roughly proportional to the inverse
square of the diameter or the pipe, and is different if the
pipe is just hanging in free space vs terminated of "flanged"
in a wall of some sort. That inertance itself provides an
acoustical reactance at the end of the tube and can itself
cause some of the energy to be reflected back down the pipe,
just like an electrical "inertance" (aka inductance).

OK,

That accounts for there being enough inertance to cause a mismatch and
retention of the sound wave at this resonance frequency, leading to a
reinforcement of that frequency wave.

I'm more interested in the conical bore effect of a clarinet, where it
apparently offers a virtual impedance against waves traveling back up the
bore; that's my take on this phenomenon. I'm not sure how high a harmonic
should be to really encounter this back-impedance.

When examining an FFT spectrum of the radiated sound level of any given
clarinet note, I have noticed that the amplitudes of the "missing" harmonics
are not zero, but only diminished to varying degrees. And I can't help but
stress that if harmonic energy is already produced by blown reed action,
said energy must be accounted for. Ergo, it does become emitted into the
space outside the clarinet bore, but perhaps does not enjoy any reverberant
buildup as occurs for the favored harmonics.

Ange