A tuning fork of frequency n is held near the | vedclass

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(C) For a tube closed at one end,the resonance lengths are given by $L = \frac{(2k-1)\lambda}{4}$,where $k = 1, 2, 3, \dots$ The first resonance occur

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$2n(L_2 - L_1)$

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(C) For a tube closed at one end,the resonance lengths are given by $L = \frac{(2k-1)\lambda}{4}$,where $k = 1, 2, 3, \dots$ The first resonance occurs at $L_1 = \frac{\lambda}{4}$. The second resonance occurs at $L_2 = \frac{3\lambda}{4}$. Subtracting the two lengths: $L_2 - L_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2}$. Therefore,the wavelength is $\lambda = 2(L_2 - L_1)$. The speed of sound $V$ is given by the formula $V = n\lambda$. Substituting the value of $\lambda$: $V = n 	imes 2(L_2 - L_1) = 2n(L_2 - L_1)$.

$A$ tuning fork of frequency $n$ is held near the open end of a tube which is closed at the other end,and the length is adjusted until resonance occurs. The first resonance occurs at length $L_1$ and the immediate next resonance occurs at length $L_2$. The speed of sound in air is

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