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(C) For an open pipe of length $L_1$,the fundamental frequency is $n_1 = \frac{v}{2 L_1}$,which implies $L_1 = \frac{v}{2 n_1}$.For a closed pipe of l

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$\frac{n_1 n_2}{n_1+2 n_2}$

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(C) For an open pipe of length $L_1$,the fundamental frequency is $n_1 = \frac{v}{2 L_1}$,which implies $L_1 = \frac{v}{2 n_1}$.For a closed pipe of length $L_2$,the fundamental frequency is $n_2 = \frac{v}{4 L_2}$,which implies $L_2 = \frac{v}{4 n_2}$.The combined pipe is closed at one end and open at the other,with total length $L = L_1 + L_2$.The fundamental frequency of the combined pipe is $n = \frac{v}{4 L} = \frac{v}{4(L_1 + L_2)}$.Substituting the values of $L_1$ and $L_2$: $n = \frac{v}{4(\frac{v}{2 n_1} + \frac{v}{4 n_2})} = \frac{v}{4v(\frac{1}{2 n_1} + \frac{1}{4 n_2})} = \frac{1}{\frac{2}{4 n_1} + \frac{1}{4 n_2}} = \frac{1}{\frac{1}{2 n_1} + \frac{1}{4 n_2}} = \frac{1}{\frac{2 n_2 + n_1}{4 n_1 n_2}} = \frac{4 n_1 n_2}{n_1 + 2 n_2}$.Wait,re-evaluating the denominator: $n = \frac{1}{\frac{2}{4 n_1} + \frac{1}{4 n_2}} = \frac{1}{\frac{1}{2 n_1} + \frac{1}{4 n_2}} = \frac{4 n_1 n_2}{2 n_2 + n_1}$.Given the options,the correct expression is $\frac{2 n_1 n_2}{n_1 + 2 n_2}$ if the length relations were different,but based on standard derivation,the result is $\frac{2 n_1 n_2}{n_1 + 2 n_2}$ is not matching. Let's re-check: $L = \frac{v}{2 n_1} + \frac{v}{4 n_2} = v \frac{2 n_2 + n_1}{4 n_1 n_2}$. Then $n = \frac{v}{4 L} = \frac{v}{4 v \frac{2 n_2 + n_1}{4 n_1 n_2}} = \frac{n_1 n_2}{n_1 + 2 n_2}$.

An open pipe of length $L_1$ has a fundamental frequency $n_1$,and a closed pipe of length $L_2$ has a fundamental frequency $n_2$. If they are joined to form a single closed pipe of length $L = L_1 + L_2$,what is the fundamental frequency of the combined pipe?

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