A closed pipe and an open pipe have their first overtone equal in frequency. Then the lengths of these pipes…
A closed pipe and an open pipe have their first overtone equal in frequency. Then the lengths of these pipes are in the ratio
$1: 2$
$2: 3$
$3: 4$
$4: 5$
Solution
The first overtone for open pipe is
$\mathrm{f}_{\mathrm{o}}=\frac{2 \mathrm{v}}{2 l_{\mathrm{o}}}$
The first overtone for closed pipe is
$\mathrm{f}_{\mathrm{c}}=\frac{3 \mathrm{v}}{4 l_{\mathrm{c}}}$
Equating the frequencies,
$\frac{2 \mathrm{v}}{2 l_{\mathrm{o}}}=\frac{3 \mathrm{v}}{4 l_{\mathrm{c}}}$
$\frac{l_{\mathrm{c}}}{l_{\mathrm{o}}}=\frac{3}{4}$