A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density $ρ$ at equilibrium position. When the cylinder is given a small downward push and released it starts oscillating vertically with a small amplitude. If the force constant of the spring is k , the frequency of oscillation of the cylinder is

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$\frac{1}{2 π} {(\frac{k + A ρ g}{A ρ g})}^{1 / 2}$

$\frac{1}{2 π} {(\frac{k - A ρ g}{M})}^{1 / 2}$

$\frac{1}{2 π} {(\frac{k + A ρ g}{M})}^{1 / 2}$

$\frac{1}{2 π} {(\frac{k + ρ g L^{2}}{M})}^{1 / 2}$

answer is B.

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Detailed Solution

When cylinder is displaced by an amount from its mean position, spring force and upthrust both will increase. hence,
Question Image
Net restoring force = extra spring force + extra upthrust
or $F = - (kx + {Axlg}^{g})$ or $a = - (\frac{kx + p Ag}{M})$ Now, $f = \frac{1}{2 π} \sqrt{| \frac{a}{x} |} = \frac{1}{2 π} \sqrt{\frac{k + ρ A g}{M}}$