A rectangular block of mass 'm' and cross-sec | vedclass

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Question

(A) When the block is in equilibrium, the weight of the block is balanced by the buoyant force: $mg = V_{submerged} \rho g = (A \cdot h) \rho g$, wher

Vedclass · Accepted Answer

$\frac{1}{2\pi} \sqrt{\frac{A\rho g}{m}}$

Vedclass · Answer

(A) When the block is in equilibrium, the weight of the block is balanced by the buoyant force: $mg = V_{submerged} \rho g = (A \cdot h) \rho g$, where '$h$' is the submerged depth.When the block is displaced vertically by a small distance '$x$',the additional buoyant force acting on it is $F_{buoyant_extra} = (A \cdot x) \rho g$.This force acts as a restoring force: $F = - (A \rho g) x$.Comparing this with the simple harmonic motion equation $F = -kx$, we get the spring constant $k = A \rho g$.The angular frequency of oscillation is $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{A \rho g}{m}}$.The frequency of oscillation is $f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{A \rho g}{m}}$.

$A$ rectangular block of mass '$m$' and cross-sectional area '$A$' floats on a liquid of density '$\rho$'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ($g$ = acceleration due to gravity).

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