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(A) At equilibrium, the weight of the block is balanced by the buoyant force:$\rho A h g = M g$where $h$ is the submerged depth.When the block is depr

Vedclass · Accepted Answer

$2 \pi \sqrt{\frac{ M }{\rho A g}}$

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(A) At equilibrium, the weight of the block is balanced by the buoyant force:$\rho A h g = M g$where $h$ is the submerged depth.When the block is depressed by a small distance $x$, the additional buoyant force acting upwards is $F_b = \rho A x g$.Since this force acts in the direction opposite to the displacement, the restoring force is $F = -\rho A x g$.Using Newton's second law, $M a = -\rho A x g$, which gives the acceleration $a = -\left(\frac{\rho A g}{M}\right) x$.Comparing this with the standard $SHM$ equation $a = -\omega^2 x$, we get $\omega^2 = \frac{\rho A g}{M}$, so $\omega = \sqrt{\frac{\rho A g}{M}}$.The period of oscillation $T$ is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{M}{\rho A g}}$.

$A$ cylindrical block of mass $M$ and area of cross-section $A$ is floating in a liquid of density $\rho$ with its axis vertical. When depressed a little and released, the block starts oscillating. The period of oscillation is . . . . . . . . . . .

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