An organ pipe $P_1$ is closed at one end and vibrating in its first overtone and another pipe $P_2$ opened…
An organ pipe $P_1$ is closed at one end and vibrating in its first overtone and another pipe $P_2$ opened at both ends vibrating in its third overtone are in resonance with a given tuning fork. Then, the ratio of lengths of $P_1$ and $P_2$ is
$\frac{1}{3}$
$\frac{2}{3}$
$\frac{8}{3}$
$\frac{3}{8}$
Solution
First overtone of closed ($P_1$) = Third overtone of open ($P_2$)
$\therefore 3\left(\frac{v}{4l_1}\right) = 4\left(\frac{v}{2l_2}\right) \Rightarrow \frac{l_1}{l_2} = \frac{3}{8}$