Solution:
For a closed pipe, \(f_c = \frac{9v}{4L_c}\). For an open pipe, \(f_o = \frac{4v}{2L_o}\). Since \(f_c = f_o\), we have \(\frac{9v}{4L_c} = \frac{4v}{2L_o} ⇒\frac{L_c}{L_o} = \frac{9}{8}\).
The \(4^{\text{th}}\) overtone of a closed organ pipe is same as that of \(3^{\text{rd}}\) overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is:
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