The displacement curves of longitudinal waves in a tube open at both ends is shown in Figs. (a) and (b). 

Let D be the diameter of the tube. We know that antinodes occur slightly outside the tube at a distance 0.3 D from the tube end.

In case of open organ pipe, the distance between two antinodes is

given by $\frac{\mathrm{\lambda }}{2}=48+2\times 0.3\mathrm{D}$          …(i)

We have, $\mathrm{\lambda }=\frac{\mathrm{v}}{\mathrm{n}}=\frac{32000}{320}=100\mathrm{cm}$

Substituting the value of $\mathrm{\lambda }$ in Eq. (i), we get

$\frac{100}{2}=48+0.6\mathrm{D}\Rightarrow 50=48+0.6\mathrm{D}\Rightarrow \mathrm{D}=\frac{10}{3}\mathrm{cm}=3.33\mathrm{cm}$

In case of closed organ pipe, 

$\frac{\mathrm{\lambda }}{4}=48+0.3\times \frac{10}{3}=49\text{ or }\mathrm{\lambda }=4\times 49=196\mathrm{cm}$

Hence the lowest frequency is, $\mathrm{f}=\frac{\mathrm{\nu }}{\mathrm{\lambda }}=\frac{32000}{196}=163.3\mathrm{Hz}$