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

(N/A) Yes,the mercury column will execute simple harmonic motion $(SHM)$.
Let the total length of the mercury column be $L$ and the cross-sectional area of the tube be $A$. Let the displacement of the mercury level from the equilibrium position along the tube be $x$.
The restoring force $F$ is due to the difference in the heights of the mercury columns in the two arms.
The difference in height $h = h_1 - h_2 = (l+x)\sin 45^{\circ} - (l-x)\sin 45^{\circ} = 2x \sin 45^{\circ} = 2x(1/\sqrt{2}) = x\sqrt{2}$.
The restoring force is $F = -mg = -(\text{mass of displaced liquid})g = -(\rho A h)g = -\rho A g (x\sqrt{2})$.
The total mass of the mercury is $M = \rho A L$.
Using Newton's second law,$F = Ma$,so $Ma = -\rho A g \sqrt{2} x$.
$(\rho A L) \frac{d^2x}{dt^2} = -\rho A g \sqrt{2} x$.
$\frac{d^2x}{dt^2} = -(\frac{g\sqrt{2}}{L})x$.
This is the equation of $SHM$ with $\omega^2 = \frac{g\sqrt{2}}{L}$.
The time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L}{g\sqrt{2}}} = 2\pi \sqrt{\frac{L}{g \sqrt{2}}}$.