A U tube of uniform bore of cross-sectional area A has been set up vertically with open ends facing up. Now m gm of a liquid of density d is poured into it. The column of liquid in this tube will oscillate with a period T such that

If the level of liquid is depressed by cm on one side, then the level of liquid in column P is 2y cm higher than B as shown.

The weight of extra liquid on the side P = 2Aydg.
This becomes the restoring force on mass M.

$\therefore \mathrm{Restoring} \mathrm{acceleration} = \frac{-2\mathrm{Aydg}}{\mathrm{M}}$

This relation satisfies the condition of SHM i.e., $\mathrm{a}\propto -\mathrm{y}$.
Hence time period

$T = 2\mathrm{\pi }\sqrt{\frac{\mathrm{Displacement}}{\mathrm{Acceleration}}}$

    = $2\mathrm{\pi }\sqrt{\frac{{\displaystyle \frac{\mathrm{y}}{2\mathrm{Aydg}}}}{\mathrm{M}}} \Rightarrow \mathrm{T} = 2\mathrm{\pi }\sqrt{\frac{\mathrm{M}}{2\mathrm{Adg}}}$