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. $\frac{1}{2\pi }\sqrt{\frac{\rho gAsin\left(\frac{{\theta }_1+{\theta }_2}{2}\right)}{m}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{\rho gA\left(sin{\theta }_1-sin{\theta }_2\right)}{m}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{\rho gA\left(sin{\theta }_1+sin{\theta }_2\right)}{m}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{\rho gAsin\left(\frac{{\theta }_1-{\theta }_2}{2}\right)}{m}}$

Answer 1

Answer: (C)

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 two arms of the tube is
$F=-\Delta PA=-\left(h_1+h_2\right)\rho gA$
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.
$F=-\left(ysin{\theta }_1+ysin{\theta }_2\right)\rho gA$
$=-\rho gA\left(sin{\theta }_1+sin{\theta }_2\right)y...\left(i\right)$
$ma=-\rho gA\left(sin{\theta }_1+sin{\theta }_2\right)y$
or $a=-\frac{\rho gA\left(sin{\theta }_1+sin{\theta }_2\right)y}{m}$
For $SHM$
$a=-{\omega }^{2}y...\left(ii\right)$
Comparing $\left(i\right)$ and $\left(ii\right)$, we get
${\omega }^2=\rho g\frac{A\left(sin{\theta }_1+sin{\theta }_1\right)}{m}$
$\omega =\sqrt{\frac{\rho gA\left(sin{\theta }_1+sin{\theta }_2\right)}{m}}$
Frequency, $\upsilon =\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{\rho gA\left(sin{\theta }_1+sin{\theta }_2\right)}{m}}$