If l1 and l2 are the lengths of air column for the first and second resonance when a tuning fork of

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If l1 and l2 are the lengths of 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:

Accepted Answer

Sol. f = $\frac{
u}{4(l_{1} + e)}$ (considering end correction)Also f = $\frac{3
u}{4(l_{2} + e)}$⇒ $\frac{4(l_{2} + e)}{3}$ = 4(l1 + e)⇒ l2 + e = 3l1 + 3e⇒ l2–3l1 =2e⇒ e = $\frac{l_{2} - 3l_{1}}{2}$

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