How are the stationary waves formed in a closed pipe ? Explain the various modes of vibrations in a closed pipe and establish the relation between their frequencies.

An organ pipe is a cylindrical tube having an air column. If one end is closed and the other is open, it is known as a closed organ pipe.
At closed end always a node and at open end an antinode will generate.
Harmonics in Closed Pipes :
i) The first harmonic in a closed pipe should have a node at the closed end and an antinode at the open end, then the length of the pipe

$\mathrm{l}=\frac{{\mathrm{\lambda }}_1}{4};{\mathrm{\lambda }}_1=4\mathrm{l}$
The fundamental frequency ${\mathrm{n}}_1=\frac{\mathrm{v}}{{\mathrm{\lambda }}_1}$
${\mathrm{n}}_1=\frac{\mathrm{v}}{4\mathrm{l}}$ .......(1)
ii) The next possible harmonic with two nodes and two antinodes it is the third harmonic, then the length of the pipe

$\mathrm{l}=\frac{3{\mathrm{\lambda }}_3}{4};{\mathrm{\lambda }}_3=\frac{4\mathrm{l}}{3}$
The frequency of the third harmonic (or) first overtone is given by ${\mathrm{n}}_3=\frac{\mathrm{v}}{{\mathrm{\lambda }}_3}$
${\mathrm{n}}_3=\frac{3\mathrm{v}}{4\mathrm{l}}$   [from equation (1)]
${\mathrm{n}}_3=3{\mathrm{n}}_1 ....\left(2\right)$
iii) Similarly the next overtone in the closed pipe is only the fifth harmonic. It will have three antinodes and three nodes. Then the length of the pipe

${\mathrm{\lambda }}_5=\frac{4\mathrm{l}}{5}$
Frequency of fifth harmonic (or) second overtone
$\begin{array}{l}{\mathrm{n}}_5=\frac{\mathrm{v}}{{\mathrm{\lambda }}_5};{\mathrm{n}}_5=\frac{\mathrm{v}}{\frac{4\mathrm{\ell }}{5}};{\mathrm{n}}_5=5\left(\frac{\mathrm{v}}{4\mathrm{\ell }}\right)\\ {\mathrm{n}}_5=5{\mathrm{n}}_1\dots ..\left(3\right)\end{array}$
from equation(1),(2)(3) the ratio of fundamental and next harmonics are 1 : 3 : 5 : ....