
$P, S$, will be $1^{\text {st }}$ and $2^{\text {nd }}$ harmonics of closed tube.
$P=\frac{v}{4 l}, S=\frac{v}{4 l}$
$Q, R$ are the $1^{\text {st }}$ and $2^{\text {nd }}$ harmonics of open pipe.
$Q=\frac{v}{2 l}, R=\frac{v}{l}$
$\therefore P: Q: R: S=1: 2: 4: 3$


