A solid cylinder of density rho_0, cross-sect | vedclass

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(B) Let the length of the submerged part of the cylinder be $h$.For the cylinder to float in equilibrium, the weight of the cylinder must be equal to

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$2\pi \sqrt {\frac{\rho_0 l}{\rho g}} $

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(B) Let the length of the submerged part of the cylinder be $h$.For the cylinder to float in equilibrium, the weight of the cylinder must be equal to the buoyant force:$Weight = Buoyant Force$$A l \rho_0 g = A h \rho g$$\Rightarrow h = \frac{\rho_0 l}{\rho}$When the cylinder is displaced downward by a small distance $x$, the additional buoyant force acting upward is $F_{net} = -(A x \rho) g$.This force acts as a restoring force, so $m a = -A \rho g x$.Since mass $m = A l \rho_0$, we have $(A l \rho_0) \frac{d^2x}{dt^2} = -A \rho g x$.$\frac{d^2x}{dt^2} = -(\frac{\rho g}{\rho_0 l}) x$.Comparing this with the standard $SHM$ equation $\frac{d^2x}{dt^2} = -\omega^2 x$, we get $\omega^2 = \frac{\rho g}{\rho_0 l}$.The time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{\rho_0 l}{\rho g}}$.

$A$ solid cylinder of density $\rho_0$, cross-section area $A$ and length $l$ floats in a liquid of density $\rho$ $(> \rho_0)$ with its axis vertical, as shown. If it is slightly displaced downward and released, the time period of oscillation will be:

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