One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

Explanation

(i) The suction pump creates the pressure difference Hence mercury rises in one limb of the U- tube .
(ii) When it is removed a net force acts one the liquid column due to the difference in level of mercury in the two limbs and therefore the liquid column executes S.H.M which can be explained as follows.

(iii) The mercury contained in a vertical U- tube upto the level A and B in its two limbs.
Suppose `rho` = density of the mercury column in both the limbs.
A is internal cross- sectional area of U- tube M is mass of mercury in U - tube M = LA`rho`
(iv) Suppose the mercury be depressed in left links to A ' by the small distance y ,m then it rises by the same amount in the right limb to position B '.
(v) Volume of mercury contained in the column of length , 2y
`Vy=Axx2y`
`m=Axx2yxxrho`
(vi) When W = weight of liquid constrained in the column of length 2 y
Then `W = mg =A xx 2y xxrho xxg`
This weight produced the restoring force f which tends to bring back the mercury to its equilibrium position.
`F=-2Ayrhog`
(vii) When a = acceleration produced in the liquid column ,then
`a=(F)/(M)=((2Arhog)y)/(LArho)`
or `a=(-2rhoyg)/(Lr(-2rhoyg)/(2Lrho)ho) " "..(1)`
`implies (y)/(a)=(2hp)/(2rhog)" " [:.L=2h]`
When h, is height of mecury in each limb Now from eqn (1) , if is clear that a `alphay` and Negative sigh shows that it acts opposite to y , so the motion of mercury in u- tube is simple harmonic in nature having time period (T) given by , `T=2pisqrt((y)/(a))=2pisqrt((2hrho)/(2rhog))`
`T=2pi sqrt((h)/(g))`