Explain the formation of stationary waves in a closed pipe and derive the equations for natural frequencies (normal modes).

(N/A) closed pipe is a pipe closed at one end and open at the other. The end in contact with the closed boundary acts as a node (displacement is zero),and the open end acts as an antinode (displacement is maximum).
Let $L$ be the length of the pipe. The boundary conditions are: displacement $y=0$ at $x=0$ (node) and displacement amplitude is maximum at $x=L$ (antinode).
The condition for an antinode at $x=L$ is $kL = (n + 1/2)\pi$,where $n = 0, 1, 2, \dots$.
Substituting $k = 2\pi / \lambda$,we get:
$\frac{2\pi}{\lambda} L = (n + 1/2)\pi$
$L = (n + 1/2) \frac{\lambda}{2} = (2n + 1) \frac{\lambda}{4}$
Thus,the possible wavelengths are $\lambda_n = \frac{4L}{2n+1}$.
Using $v = f\lambda$,the natural frequencies (normal modes) are:
$f_n = \frac{v}{\lambda_n} = (2n + 1) \frac{v}{4L}$,where $n = 0, 1, 2, \dots$.
For $n=0$,the fundamental frequency is $f_0 = \frac{v}{4L}$.
The higher frequencies are odd harmonics of the fundamental frequency: $3f_0, 5f_0, 7f_0, \dots$.