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We were discussing a problem on the Resonance Tube Experiment in class when my teacher wrote down the following equations:

My teacher's figure

AN and N stand for the positions of the antinode and node, e stands for the end correction and $l_1$, $l_2$ and $l_3$ are lengths of the tube in the three cases. What I am wondering is, why can't the end correction be applied inside the tube? Mathematically. why can't e be negative? Does this have anything to do with the reason for end correction?

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Here is the reason that e is always positive, coming out of the open end:

For an open-ended tube to radiate sound at its resonant frequency requires that the air inside the tube be coupled to the air outside it. That coupling adds a little mass to the air mass inside the tube- in essence, it increases the effective length of the tube slightly, causing it to play slightly flat.

Note that as the tube radiates, power is leaving the oscillating tube and disappears into the air outside the tube. This power loss looks just like a damping term in the equations of motion for the oscillator- and as you already know, the damped resonant frequency is always slightly lower than the undamped resonant frequency.

But now you know the physics behind why that is true!

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  • $\begingroup$ I am still pretty uncertain about the 'mass being added'. It would be great if you explained that part a little more in detail if possible. $\endgroup$ Jul 13, 2021 at 15:53
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The, e, assumes that the anti-node lies beyond the end of the open tube. Combining two of the resonance equations permits finding, λ, and, e; but you may find that e changes as, λ, gets smaller.

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  • $\begingroup$ Is it physically possible for the organ pipe to exceed the length of the antinode? Why or why not? $\endgroup$ Mar 31, 2021 at 5:57

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