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Question

(A) When the block is displaced vertically by a small distance $x$ from its equilibrium position,the additional volume of liquid displaced is $V = A x

Vedclass · Accepted Answer

$n \propto \sqrt{A}$

Vedclass · Answer

(A) When the block is displaced vertically by a small distance $x$ from its equilibrium position,the additional volume of liquid displaced is $V = A x$.The additional buoyant force acting on the block is $F = ho V g = ho (A x) g$.This force acts as a restoring force,so $F_{	ext{restoring}} = -ho A g x$.According to Newton's second law,$F = M a$,where $M$ is the mass of the block.Thus,$M a = -ho A g x$,which gives $a = -(\frac{ho A g}{M}) x$.Comparing this with the standard $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = \frac{ho A g}{M}$,so $\omega = \sqrt{\frac{ho A g}{M}}$.Since frequency $n = \frac{\omega}{2 \pi}$,we have $n = \frac{1}{2 \pi} \sqrt{\frac{ho A g}{M}}$.Therefore,$n \propto \sqrt{A}$.

$A$ 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 $n$. Then:

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