If l_1 and l_2 are the lengths of the air col | vedclass

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(C) In a resonance tube,the length of the air column for the first resonance is $l_1 = \frac{\lambda}{4} - e$,where $e$ is the end correction.For the

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$\frac{l_2 - 3l_1}{2}$

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(C) In a resonance tube,the length of the air column for the first resonance is $l_1 = \frac{\lambda}{4} - e$,where $e$ is the end correction.For the second resonance,the length is $l_2 = \frac{3\lambda}{4} - e$.Subtracting the two equations: $l_2 - l_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2}$.Thus,$\lambda = 2(l_2 - l_1)$.Substituting $\lambda$ into the first resonance equation: $l_1 = \frac{2(l_2 - l_1)}{4} - e = \frac{l_2 - l_1}{2} - e$.Therefore,the end correction $e = \frac{l_2 - l_1}{2} - l_1 = \frac{l_2 - 3l_1}{2}$.The displacement antinode is located at a distance $e$ above the open end of the tube. Hence,the distance is $\frac{l_2 - 3l_1}{2}$.

If $l_1$ and $l_2$ are the lengths of the air column for the first and second resonance when a tuning fork of frequency $n$ is sounded on a resonance tube,then the distance of the displacement antinode from the top end of the resonance tube is:

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