The fundamental frequency of cylindrical open tube

$n=\frac{v}{2L}=390\mathrm{ }\mathrm{H}\mathrm{z}$

When it is immersed in water it becomes a closed tube of length, ${\frac{3}{4}}^{\mathrm{t}\mathrm{h}}$ of the initial length.

Therefore, its fundamental frequency is

$n^{\mathrm{\prime }}=\frac{v}{4\left(\frac{3}{4}L\right)}=\frac{v}{3L}=\frac{2}{3}\left(\frac{v}{2L}\right)$

$=\frac{2}{3}\times 390 \mathrm{H}\mathrm{z}=260 \mathrm{H}\mathrm{z}$