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(C) The displacement is given as $y = A \cos(30^{\circ})$.Given amplitude $A = 40 \, cm = 0.4 \, m$.At the given time,the displacement $y = 40 \cos(30

Vedclass · Accepted Answer

$4$

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(C) The displacement is given as $y = A \cos(30^{\circ})$.Given amplitude $A = 40 \, cm = 0.4 \, m$.At the given time,the displacement $y = 40 \cos(30^{\circ}) = 40 	imes \frac{\sqrt{3}}{2} = 20\sqrt{3} \, cm = 0.2\sqrt{3} \, m$.The kinetic energy of a simple harmonic oscillator is given by $K.E. = \frac{1}{2} k(A^2 - y^2)$.Substituting the given values: $200 = \frac{1}{2} k((0.4)^2 - (0.2\sqrt{3})^2)$.$200 = \frac{1}{2} k(0.16 - 0.12)$.$200 = \frac{1}{2} k(0.04)$.$200 = k(0.02)$.$k = \frac{200}{0.02} = 10000 = 1.0 	imes 10^4 \, Nm^{-1}$.Comparing this with $1.0 	imes 10^x \, Nm^{-1}$,we get $x = 4$.

Column-$I$	Column-$II$
$(a)$ $y = \frac{A}{\sqrt{2}}$	$(i)$ $K = \frac{3E}{4}$
$(b)$ $y = \frac{\sqrt{3}A}{2}$	$(ii)$ $K = \frac{E}{4}$
	$(iii)$ $K = \frac{E}{2}$

At a given point of time,the displacement of a simple harmonic oscillator is given by $y = A \cos(30^{\circ})$. If the amplitude is $40 \, cm$ and the kinetic energy at that time is $200 \, J$,the value of the force constant is $1.0 \times 10^{x} \, Nm^{-1}$. The value of $x$ is ......

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