A wooden cube (density of wood d) of side l f | vedclass

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(B) Let the cube be pushed down by a small displacement $x$. The additional buoyant force acting on the cube is $F = -A \rho g x$,where $A = l^2$ is t

Vedclass · Accepted Answer

$2\pi \sqrt {\frac{{ld }}{{\rho g}}} $

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(B) Let the cube be pushed down by a small displacement $x$. The additional buoyant force acting on the cube is $F = -A \rho g x$,where $A = l^2$ is the area of the cross-section.This force acts as a restoring force,so $F = -k x$,where $k = A \rho g = l^2 \rho g$.The mass of the cube is $m = l^3 d$.The time period of simple harmonic motion is given by $T = 2\pi \sqrt{\frac{m}{k}}$.Substituting the values,we get $T = 2\pi \sqrt{\frac{l^3 d}{l^2 \rho g}}$.Simplifying this,we obtain $T = 2\pi \sqrt{\frac{ld}{\rho g}}$.

$A$ wooden cube (density of wood $d$) of side $l$ floats in a liquid of density $\rho$ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released,it performs simple harmonic motion of period $T$. Then,$T$ is equal to

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