The air column in a pipe closed at one end is vibrating in its second overtone. The end correction may be neglected. Let $P_0$ denote the pressure amplitude at the positions of the displacement nodes. Then the pressure amplitude at the middle of the air column is:

The figure shows the pressure standing wave pattern. The closed end is a position of pressure antinode and the free end is a position of pressure node. The air column is vibrating in its second overtone (5th harmonic), hence there are two and a half loops. If I is the length of the tube, then

                 $\frac{5\lambda }{4}=I\Rightarrow \lambda =\frac{4l}{5}$

Hence $k=\frac{2\pi }{\lambda }=\frac{5\pi }{2l}$

If we choose the x-axis and the origin, as shown in the figure, then the pressure amplitude as a function of x will be

$P\left(x\right)=P_0|\sin (kx\left)\right|=P_0\left|\sin \left(\frac{5\pi }{21}x\right)\right|$

At the middle of the tube $(x=l/2)$ , the pressure amplitude is $P_0\left|\sin \left(\frac{5\pi }{4}\right)\right|=\frac{P_0}{\sqrt{2}}$