Solution :
Suppose the liquid in left side limb is displaced slightly by $$y,$$ the liquid in right limb will increase by $$\frac{y}{2}.$$ The restoring force
$$\eqalign{
& F = - PA = - \rho g\left( {\frac{{3y}}{2}} \right) \times 2A = 3\rho gA\left( { - y} \right). \cr
& a = \frac{F}{m} = \frac{{3\rho gA\left( { - y} \right)}}{m} \cr} $$

On comparing with, $$a = - {\omega ^2}y,$$ we get
$$\omega = \sqrt {\frac{{3\rho gA}}{m}} \,\,{\text{and}}\,\,T = 2\pi \sqrt {\frac{m}{{3\rho gA}}} $$