XPost: sci.physics.acoustics, sci.physics   
   From: a.campanella@att.net   
      
   "Dick Pierce"  wrote in message   
   news:ifak2g$nj2$1@speranza.aioe.org...   
      
   > 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   
      
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