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Q.
$n_1$ is the frequency of the pipe closed at one end and $n_2$ is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe so formed
Frequency of closed pipe, $n_1=\frac{v}{4l_1} \Rightarrow l_1 = \frac{v}{4n_1}$
Frequency of open pipe, $n_2=\frac{v}{2l_1} \Rightarrow l_2 = \frac{v}{2n_2}$
when both pipe are joined then length of closed pipe
$ l = l_1 + l_2 $
$ \frac{v}{4n} =\frac{v}{4n_1}= \frac{v}{2n_2}$
or $ \frac{1}{2n} =\frac{1}{2n_1}= \frac{1}{n_2}$
or $ \frac{1}{2n} =\frac{n_2 + 2n_1}{2n_1n_2}$
or $ n =\frac{n_1n_2}{n_2 + 2n_1}$