Wave Mechanics Question 594
Question: A U-tube is of non uniform cross-section. The area of cross-sections of two sides of tube are A and 2A (see fig.). It contains non-viscous liquid of mass m. The liquid is displaced slightly and free to oscillate. Its time period of oscillations is
Options:
A) $ T=2\pi \sqrt{\frac{m}{3\rho gA}} $
B) $ T=2\pi \sqrt{\frac{m}{2\rho gA}} $
C) $ T=2\pi \sqrt{\frac{m}{\rho gA}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Suppose the liquid in left side limb is displaced slightly by y, the liquid in right limb will increase by $ y/2 $ . The restoring force $ F=-PA=-\rho g( \frac{3y}{2} )\times 2A=3\rho gA(-y) $ . $ a=\frac{F}{m}=3\rho gA(-y)/m $ On comparing with, $ a=-,{{\omega }^{2}}y $ , we get $ \omega =\sqrt{\frac{3\rho gA}{m}} $ and $ T=2\pi \sqrt{\frac{m}{3\rho gA}} $
BETA
