Question

A vertical U – tube of uniform cross – sectional area A contains a liquid of density $\rho$ The total length of the liquid column in the tube is L. the liquid column is disturbed by gently blowing into the tube. If viscous effects are neglected, the time period of the resulting oscillation of the liquid column is given by

• A. $T=2\pi \sqrt{\frac{L\rho }{gA}}$
• B. $T=2\pi \sqrt{\frac{L}{\rho gA}}$
• C. $T=2\pi \sqrt{\frac{L}{g}}$
• D. $T=2\pi \sqrt{\frac{L}{2g}}$

Answer 1

The correct option is D $T=2\pi \sqrt{\frac{L}{2g}}$

Let the liquid level in the left arm be depressed by, say, x. The liquid level in the right arm is raised by an equal amount x. When left, the liquid levels will oscillate in each arm about their respective equilibrium position A, A’. If viscous effects are neglected, the liquid will never come too rest. Since the column in the right arm is higher by 2x than the column in the left arm,the mass of this column of liquid is m = 2Apx. The gravitational force mg of this column provides the restoringforce F. Thus
F = −mg =−2Apgx = −kx
Where k = 2Apg.

If L is the total length of the liquid in the U-tube, the total mass M of the oscillating liquid is
M = pAL
Hence the acceleration of the liquid column is given by
Acceleration $a=\frac{force}{mass}$
=$\frac{F}{m}=-\frac{kx}{M}$
=$\frac{2Apg}{pAL}\times$
=$-\frac{2g}{L}\times$
$Or,a=-{\omega }^2\times with\omega =\sqrt{\frac{2g}{L}}$
Hence the motion is simple harmonic. The time period of the motion is given by
$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{L}{2g}}$
Note: The period of oscillation is independent of the density p of the liquid and the cross-sectional area A of the U-tube.