A cylindrical tube open at both ends possesses a fundamental...

For an open tube of length $L$, the fundamental frequency is given by

$\nu = \dfrac{v}{2L}$

When the tube is immersed vertically in water such that half of its length is submerged, it behaves as a closed tube with an air column of length $L' = \dfrac{L}{2}$.

The fundamental frequency of a closed tube is given by

$\nu' = \dfrac{v}{4L'}$

Substituting $L' = \dfrac{L}{2}$, we get

$\nu' = \dfrac{v}{4 \left( \dfrac{L}{2} \right)} = \dfrac{v}{2L}$

Comparing with the initial frequency, we have

$\nu' = \nu$