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Suppose the water on one side tube is pressed down by a a height x meters using a smooth frictionless piston for example.  Then water in the other tube will rise by y meters from its original equilibrium position.  Let A be the area of cross section of the tube.  let ρ be the mass volume density of water.

The difference between the water levels in the two tubes will be  2 y  meters.
the pressure experienced by the piston will then be =  ρ g (2 y)
The restoration force  that water exerts on the piston =  2 ρ g y A

We see that  the restoration force = mass * acceleration
         = m d²y/ dt² of the water particles on the surface just below the piston
         =  - 2 ρ A g y
   The minus sign is because the restoration force acts in the direction opposite to positive x direction.  Clearly this  is an equation for the displacement y being in SHM.
   The solution for this differential equation is that:
               ω² =  2 ρ A g
               ω = √(2 ρ A g)

     Time period of SHM =  2π/ω  = 2π / [√2ρ A g]

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