A light hollow cube of side length 10 cm and | vedclass

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Question

(A) When the cube is pushed down by a small displacement $x$,the additional buoyant force acting on it is $F_b = A \rho g x$,where $A = L^2$ is the ar

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) When the cube is pushed down by a small displacement $x$,the additional buoyant force acting on it is $F_b = A ho g x$,where $A = L^2$ is the area of the base of the cube.Since the cube is light and floating,the restoring force is $F = -L^2 ho g x$.Using Newton's second law,$m \frac{d^2x}{dt^2} = -L^2 ho g x$,which gives $\frac{d^2x}{dt^2} = -(\frac{L^2 ho g}{m}) x$.This is the equation of simple harmonic motion with angular frequency $\omega = \sqrt{\frac{L^2 ho g}{m}}$.The time period is $T = 2 \pi \sqrt{\frac{m}{L^2 ho g}}$.Given $m = 10 \ g = 10^{-2} \ kg$,$L = 10 \ cm = 0.1 \ m$,$ho = 10^3 \ kg/m^3$,and $g = 10 \ m/s^2$.Substituting the values: $T = 2 \pi \sqrt{\frac{10^{-2}}{(0.1)^2 	imes 10^3 	imes 10}} = 2 \pi \sqrt{\frac{10^{-2}}{0.01 	imes 10^4}} = 2 \pi \sqrt{\frac{10^{-2}}{10^2}} = 2 \pi \sqrt{10^{-4}} = 2 \pi 	imes 10^{-2} \ s$.Comparing this with $y \pi 	imes 10^{-2} \ s$,we get $y = 2$.

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