The suction pump creates the pressure difference, thus mercury rises in one limb of the U-Tube. When it is removed, a net force acts on the liquid column due to the difference in levels of mercury in the two limbs and hence the liquid column executes S.H.M. which can be explained as :Consider the mercury contained in a vertical U-tube upto the level P and Q in its two limbs.Let $\rho =$density of the mercury.$L =$ total length of the mercury column in both the limbs.$A =$ internal cross-sectional area of U-tube.$m =$ mass of mercury in U-tube $= LA\rho$.Let the mercury be depressed in left limb to $P'$ by small distance y, then it rises by the same amount in the right limb to position $Q'$∴ Difference in levels in the two limbs $= P'Q' = 2y$∴ Volume of mercury contained in the column of length $2y = A \times 2y$$m' = A \times 2y \times \rho$If W = weight of liquid contained in the column of length $2y$Then $W = m' g = A \times 2y \times \rho \times g$This weight produces the restoring force $(F)$ which tends to bring back the mercury to its equilibrium position. $\therefore F=-2 A y \rho g=-(2 A \rho g) y$If $a =$ acceleration produced in the liquid column, then$a= \frac{F}{m}= - \frac{(2 A \rho g) y}{L A \rho}$$= - \frac{2 y g}{L}$$= - \frac{2g}{2h} y (\because L=2h)\dots(i)$where $h =$ height of mercury in each limb. Now from (i), it is clear that $a \propto y$ and –ve sign shows that it acts opposite to $y$, so the motion of mercury in U-tube is simple harmonic in nature having time period (T) given by$T=2\pi\sqrt{\frac{y}{a}}$$=2\pi\sqrt{\frac{2h}{2g}}=2\pi\sqrt{\frac{h}{g}}$