Solution:
The first overtone of a closed pipe of length \(L_1\) has frequency \(f_{c,1} = \frac{3v}{4L_1}\) and that of an open pipe of length \(L_2\) is \(f_{o,1} = \frac{v}{L_2}\). Equating the two frequencies gives \(\frac{3v}{4L_1} = \frac{v}{L_2}\), which simplifies to \(\frac{L_1}{L_2} = \frac{3}{4}\).
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