The lengths of two open organ pipes are l and | vedclass

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(C) For an open organ pipe of length $L$,the fundamental frequency is given by $n = \frac{v}{2L}$.For the first pipe of length $l$,the frequency is $n

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$\frac{v \Delta l}{2l^2}$

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(C) For an open organ pipe of length $L$,the fundamental frequency is given by $n = \frac{v}{2L}$.For the first pipe of length $l$,the frequency is $n_1 = \frac{v}{2l}$.For the second pipe of length $(l + \Delta l)$,the frequency is $n_2 = \frac{v}{2(l + \Delta l)}$.The beat frequency is the difference between the two frequencies: $f_{beat} = n_1 - n_2 = \frac{v}{2l} - \frac{v}{2(l + \Delta l)}$.Taking $\frac{v}{2}$ as a common factor: $f_{beat} = \frac{v}{2} \left( \frac{1}{l} - \frac{1}{l + \Delta l} \right)$.Simplifying the expression: $f_{beat} = \frac{v}{2} \left( \frac{l + \Delta l - l}{l(l + \Delta l)} \right) = \frac{v \Delta l}{2l(l + \Delta l)}$.Since $\Delta l$ is very small compared to $l$,we can approximate $l + \Delta l \approx l$.Therefore,$f_{beat} \approx \frac{v \Delta l}{2l^2}$.

The lengths of two open organ pipes are $l$ and $(l + \Delta l)$ respectively. Neglecting end correction,the frequency of beats between them will be approximately (Here $v$ is the speed of sound).

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