Question

A uniform U tube kept vertical contains a very slightly mobile liquid of length l. If the liquid in one limb is depressed by x and then released, find the period of oscillation of the liquid.

Answer 1

Hint:-Since the position of the liquid in the LHS is lowered by the original position and the position of the liquid is increased on the RHS by the same distance the LHS was lowered. So, the difference between the two positions will be 2x. As there is a decrease of -x on LHS and an increase of +x on the RHS. Apply the formula for pressure and then find the time period by using the formula. $T = \dfrac{{2\pi }}{\omega }$. Where T = Time period;$\omega $ = angular velocity.

Complete step-by-step solution:-
The formula for finding the pressure is:
$P = \rho gh$;
Where,
P = Pressure.
g = gravitational acceleration.
h = height.
$P = \dfrac{F}{A}$;
Where;
P = Pressure;
F = Force;
A = Area;
Complete step by step solution: Find the pressure.
$P = \rho gh$;
Put in the given values.
$P = \rho g2x$;
Write the general formula for pressure
$P = \dfrac{F}{A}$ ;
Find the Force,
$F = P \times A$;
As a restoring force the sign would be negative
$F = - \rho g2x \times A$;
According to Newton’s Second law
$F = ma$;
Put the above relation in to the equation.$F = \rho g2x \times A$;
$ma = - \rho g2x \times A$;
Find acceleration:
$a = - \dfrac{{\rho g2x \times A}}{m}$;
For Simple Harmonic Motion (SHM) the acceleration is given by
$a = - {\omega ^2}x$;
Equate the two relations of acceleration,
 $ - {\omega ^2}x = - \dfrac{{\rho g2x \times A}}{m}$
${\omega ^2} = \dfrac{{2\rho gA}}{m}$
Simplify,
$\omega = \sqrt {\dfrac{{2\rho gA}}{m}} $
The time period would be
$T = \dfrac{{2\pi }}{\omega }$
Put the value of $\omega $ in the above equation
$T = \dfrac{{2\pi }}{{\sqrt {\dfrac{{2\rho gA}}{m}} }}$
$T = 2\pi \times \sqrt {\dfrac{m}{{2\rho gA}}} $
Final Answer: The period of oscillation of the liquid is $T = 2\pi \times \sqrt {\dfrac{m}{{2\rho gA}}} $.

Note:- Here in this question we first have to find out the pressure, then form a relation between force and pressure, after that apply Newton’s Second law and find the acceleration. Then equate the calculated acceleration with acceleration for SHM and find the angular velocity. Apply the formula for time period for SHM and put the value of angular velocity.