One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45° each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

In this problem we prove the S.H.M. by deriving standard equation ------(1),

which represents simple harmonic equation.

Let the height of the fluid in the left and right column are respectively.

Consider an element with mass which is at a height of unit from the bottom in the left column can be related as follows,

, where ρ = density of the fluid in the tube

A=Area of cross-section of the tube

For total potential energy we integrate the above,

Similarly, for right column we have,

∴ total potential energy,

Now initially

Let the vertical height of the liquid on both the tube be and when the pressure difference is produced the liquid in one tube drops by and liquid in the other raises by height of .

New height at left column for the fluid =

And new height at left column for the fluid =

∴ the total

Now the change in potential energy

Change in kinetic energy, ----(2)

Now if is the length of the tube,

then

And mass m is given as

Putting the above in the equation (2), we get,

∴ total energy =

Now the change in total energy is =0

∴

Differentiating w.r.t. time t,

()

(∵ acceleration and here )

Now putting , we get,

-------(3)

Comparing (1) and (3) we see that the above equation represents S.H.M. and on comparing we get,