$L=\frac{n \lambda}{2}$

where, $\lambda$ is the wavelength of wave and $n$ is an integer and by putting $n=1,2,3……….$

we get the modes of vibration.

n=1 gives first harmonics, $n=2$ gives second harmonics and so on.

Here, an open organ pipe of length $L$ vibrates in second harmonic mode,

hence the length of pipe is

$L=\frac{2 \lambda}{2}$

$L=\lambda$

Hence, the length of pipe is equal to the distance between two consecutive antinodes

at both open ends. The nodes are at distance $\frac{\lambda}{4}$ from the antinodes. since there is no

displacement of particles at nodes the pressure vibration is maximum there.

Hence, in open organ pipe the pressure vibration is maximum at a distance

$\frac{L}{4}(\text { since }, L=\lambda)$ from either end inside the tube as shown in figure.