Question

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 horizontal at an angle of $45^\circ $ each. A small pressure difference is created between 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?

Answer 1

Hint: Simple harmonic motion is a special type of periodic motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of this restoring force is always directed to the mean position.

Complete step by step solution:
Let’s assume that the length of the liquid column in each arm of the v-tube is in equilibrium position to be $'l'$. Also let $'\rho '$ be the density of the mercury in the v- tube and $'A'$ be the area of the cross section of the liquid column.
If the distance pushed by mercury in one arm of the v-tube is $'x'$ then the level of mercury in the other arm will be raised by $'x'$.
The pressure difference experienced due to the difference in level of mercury in v-tube will be;
$\Delta P = 2x(\sin 45^\circ )\rho g$
The thrust perpendicular to the liquid meniscus in the v-tube is;
$F = - \dfrac{{(\Delta P)A}}{{\cos 45^\circ }} = - 2\rho gAx - - - - - (1)$
From (1) we can conclude that
$F = - kx$
$F \propto ( - x)$
Which shows that the motion is in simple harmonic motion(shm)
So, $k = 2\rho gA$
Time period $T = 2\pi \sqrt {\frac{m}{k}} = 2\pi \sqrt {\dfrac{{2lA\rho }}{{2\rho gA}}} = 2\pi \sqrt {\dfrac{l}{g}} $
$\therefore $Time period of oscillation$ = 2\pi \sqrt {\dfrac{l}{g}} $

Note:
All the simple harmonic motions are oscillatory and periodic but all oscillatory motions are not simple harmonic. Simple harmonic motion is also a type of oscillatory motion where the acceleration of the particle at any position is always directly proportional to the displacement from the mean position