A uniform cylinder of length L and mass M hav | vedclass

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Question

(D) Let $x$ be the downward displacement of the cylinder from its equilibrium position. When the cylinder is displaced by $x$,the additional buoyant f

Vedclass · Accepted Answer

$2\pi \left[ \frac{M}{k + A\sigma g} \right]^{1/2}$

Vedclass · Answer

(D) Let $x$ be the downward displacement of the cylinder from its equilibrium position. When the cylinder is displaced by $x$,the additional buoyant force acting upwards is $F_b = A \sigma g x$. The spring force acting upwards is $F_s = kx$. The net restoring force is $F_{net} = -(k + A \sigma g)x$. Comparing this with the standard $SHM$ equation $F = -m \omega^2 x$,we get $m \omega^2 = k + A \sigma g$. Thus,the angular frequency is $\omega = \sqrt{\frac{k + A \sigma g}{M}}$. The time period is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{M}{k + A \sigma g}}$. Since the cylinder is only half-submerged,the buoyant force is constant as long as it remains partially submerged. The derived formula is exact for small oscillations.

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