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B Why does end correction in pipes change with radius

  1. Apr 29, 2017 #1
    This being WRT resonating pipes. Apparently the acoustical length of the pipe is different to the physical length due to the vibration of the sound particles moving the particles at the opening so that the physical length is no longer the length of resonance.

    I've found many sources on the broad topic on end correction, and I know that it's commonly accepted that the end correction of a closed pipe is ~0.6*r, where r is the radius of the pipe. I've been doing some research, and I really can't understand why this is the case, however, no matter where I search, I can't find any sources on the topic.

    P.S any sources would be greatly welcomed!
     
  2. jcsd
  3. Apr 29, 2017 #2
    That makes some sense, but I'm still a little confused... Say I have the following closed pipe resonating at it's first harmonic:
    _______________..D
    |A..................B....C
    |A..................B....C
    |A____________B....C
    .......................D
    Then, as I understand it, the air particles at points A aren't moving, or they aren't moving much, and the air particles at points B are moving the most. Will those particles be moving laterally to points C? Or will they also move to points D? When they move to point D, do they move back into the pipe? If so, than that means that those particles will have traveled over a greater area with a pipe with a greater radius, yes, but I don't see how that actually increases the distance that the particles are traveling. Wouldn't that distance just be affected by the energy of the particles B as they're being displaced? It is the extra traveling distance of the vibrating particles outside of the length of the pipe which creates end correction, no?
     
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