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According to Young and Freedman's Physics textbook, in open-ended air columns like some woodwind instruments, the position of the displacement antinode extends a tiny amount beyond the end of the column.

UCONN's website states that the end correction for a cylinder could be found by:

$$d=0.6 r$$ where $d$ is the end correction, the distance which the antinode extends beyond the end of the pipe and $r$ is the radius of the cylindrical pipe.

How does one derive that relationship?

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  • $\begingroup$ Note that some relationships in physics are "derived" through fitting experimental data. I don't know that it's the case here, but be aware that it does happen this way too. $\endgroup$
    – Kyle Kanos
    Commented Oct 24, 2015 at 14:48
  • $\begingroup$ You can check this if you like...... physics.stackexchange.com/q/209335 $\endgroup$ Commented Oct 24, 2015 at 15:38

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This is actually a fairly involved calculation that was done by Levine and Schwinger in 1948. If you are interested , the reference is H.Levine and J. Schwinger, "On the radiation of sound from an unflanged circular pipe", Physical Review 73:383-406

I'll not attempt to replicate that calculation here but will try to describe the main points.

The main factor leading to the end correction is the boundary condition at the end of the pipe. The continuity of air pressure and velocity at the end of the pipe requires that the mechanical impedance of the wave equal the acoustic radiation impedance of the end of the pipe. So, the acoustic radiation impedance at the end of the pipe determines the end correction on the antinode. The radiation impedance is not zero, but is the radiation impedance of the pipe end.

To calculate the radiation impedance of the pipe end, it's treated as an unflanged piston embedded in a plane. Also it's assumed that the wavelength of sound is much larger than the diameter of the pipe. The resulting radiation impedance has real and imaginary parts, and it's the imaginary part that leads to the end correction. Levine and Schwinger arrived at a value of 0.6133d for the end correction to the effective length of the pipe.

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    $\begingroup$ "The resulting radiation impedance has real and imaginary parts, and it's the imaginary part that leads to the end correction". Could you please elaborate on this? Why does the imaginary part of the impedance lead to the end correction? $\endgroup$
    – mike
    Commented Sep 28, 2017 at 5:22

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