A U tube of uniform cross sectional area contains mercury upto height h in either limb. The mercury in one limb is depressed a little and then released. Obtain an expression for the time period of oscillation.

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Solution

$L e t x b e t h e d i s p a l a c e m e n t o f m e r c u r y t u b e, t h e n F o e c e a c t i n g u p w a r d, F = - A x ρ g N o w m a = - A x ρ g a = - \frac{A x ρ g}{m} = ω^{2} x T i m e p e r i o d, T = \frac{2 π}{ω} = 2 π \sqrt{\frac{m}{A ρ g}} T = 2 π \sqrt{\frac{m}{π r^{2} ρ g}} = \frac{2}{r} \sqrt{\frac{π m}{ρ g}}$

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