Solution:
For the fifth overtone of a closed pipe, \(f_c = 11 \left(\frac{v}{4L_c}\right)\). For the third overtone of an open pipe, \(f_o = 4 \left(\frac{v}{2L_o}\right)\). Equating \(f_c = f_o\) yields \(\frac{L_c}{L_o} = \frac{11}{8}\).
The fifth overtone of a closed pipe is observed to be unison with third overtone of an open pipe. The ratio of the lengths of the pipes is
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