One end of a $U$-tube containing mercury is connected to a suction pump and the other end to the 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.

(N/A) Let the area of cross-section of the $U$-tube be $A$ and the density of mercury be $\rho$.
When the mercury is displaced by a small distance $h$ in one arm, the level in the other arm also changes by $h$, creating a total height difference of $2h$.
The restoring force $F$ is equal to the weight of the excess mercury column of height $2h$.
$F = -(\text{Volume} \times \text{Density} \times g) = -(A \times 2h \times \rho \times g) = -2A\rho gh$.
This force is proportional to the displacement $h$, i.e., $F = -kh$, where $k = 2A\rho g$ is the force constant.
Since the restoring force is directly proportional to the displacement and directed towards the equilibrium position, the motion is simple harmonic.
The mass of the mercury column $m$ is given by $m = \text{Volume} \times \text{Density} = A \times l \times \rho$, where $l$ is the total length of the mercury column.
The time period $T$ is given by $T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{Al\rho}{2A\rho g}} = 2\pi \sqrt{\frac{l}{2g}}$.
Thus, the mercury column executes simple harmonic motion.