Question

A non viscous liquid of density $\rho$ is filled in a tube with A as the area of cross section, as shown in the figure. If the liquid is slightly depressed in one of the arms, the liquid column oscillates with a frequency

• A. $\frac{1}{2\pi}\sqrt{\frac{\rho g\text{Asin}\left( \frac{\theta_{1} + \theta_{2}}{2} \right)}{m}}$
• B. $\frac{1}{2\pi}\sqrt{\frac{\rho gA(\text{sin}\theta_{1} - \sin\theta_{2})}{m}}$
• C. $\frac{1}{2\pi}\sqrt{\frac{\rho gA(\text{sin}\theta_{1} + \sin\theta_{2})}{m}}$
• D. $\frac{1}{2\pi}\sqrt{\frac{\rho g\text{Asin}\left( \frac{\theta_{1} - \theta_{2}}{2} \right)}{m}}$

Answer 1

Answer: (C)

$\frac{1}{2\pi}\sqrt{\frac{\rho gA(\text{sin}\theta_{1} + \sin\theta_{2})}{m}}$

Answer 2

Initially the level of liquid in the two limbs will be at the same height if the liquid is depressed by y in one limb, it will rise by y along the length of the tube in the other limb the force that is responsible for restoring the liquid levels in the tow arms of the tube is

$F = - \Delta PA = - (h_{1} + h_{2})\rho gA,$

Where $\Delta P$ is the pressure difference and A is the area of cross section of the tube $h_{1}$ and $h_{2}$ being the rise and fall of liquid levels in the two arms in vertical direction respectively.

$F = - (y\sin\theta_{1} + y\sin\theta_{2})\rho gA,$

$= - \rho gA(\sin\theta_{1} + \sin\theta_{2})y$ …… (i)

$ma = - \rho gA(\sin\theta_{1} + \sin\theta_{2})y$

Or $a = \frac{\rho ga(\sin\theta_{1} + \sin\theta_{2})y}{m}$

For SHM

$a = \omega^{2}y$ …… (ii)

Comparing (i) and (ii) we get

$\omega^{2} = \frac{\rho gA(\sin\theta_{1} + \sin\theta_{2})}{m}$

$\omega = \sqrt{\frac{\rho gA(\sin\theta_{1} + \sin\theta_{2})}{m}}$

Frequency $\upsilon = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{\rho gA(\sin\theta_{1} + \sin\theta_{2})}{m}}$