Question

A vertical U tube of uniform cross section contains water to a heightof 0.3 m. Show that, if water in one of the limbs is depressed andthen released, the oscillations of the water column in the tube areSHM. Calculate its time period also.

Answer 1

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]