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Calculate th eperiod of small oscillations of a hydrometer which was slightly pushed down in the vertical direction. The mass of the hydrometer is `m=50g`, the radius of its tube is `r=3.2mm`, the density of the liquuid is `rho=1.00 g//cm^(3)` . The rsistance of the liquid is assumed to be negligible. |
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Answer» If the hydromoter were in equlibrium or floacting, its weight will be balanced by the buoyancey force acting on it by the fluid. During its small oscillation, let us locate the hydrometer when it is at a vertically downward distance `x` from its equilibrium position. Obviously teh net unbalanced force on the hydrometer is the excess buoyancy force directed upward and equals `pir^(2)x rho g`. Hence for the hydrometer. `m ddot(x)=-pir^(2)rho g x` or, `ddot(x) =-(pi r^(2) rho g )/(m)x` Hence the sought time period `T=2alpha sqrt((m)/(pir^(2)rho g))=2.5 s.` |
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