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(A) Given,period $T = 6 \ s$. The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$ and total energy $(T.E.)$ is $T.E. =

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$0.75$

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(A) Given,period $T = 6 \ s$. The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2} m \omega^2 (A^2 - x^2)$ and total energy $(T.E.)$ is $T.E. = \frac{1}{2} m \omega^2 A^2$. We are given that $K.E. = 50\% 	ext{ of } T.E.$,so $K.E. = \frac{1}{2} T.E.$. Substituting the expressions: $\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} (\frac{1}{2} m \omega^2 A^2)$. This simplifies to $A^2 - x^2 = \frac{A^2}{2}$,which gives $x^2 = \frac{A^2}{2}$ or $x = \frac{A}{\sqrt{2}}$. Since the particle starts from the mean position,the displacement equation is $x = A \sin(\omega t)$. Substituting $x = \frac{A}{\sqrt{2}}$,we get $\frac{A}{\sqrt{2}} = A \sin(\frac{2\pi}{T} t)$. $\frac{1}{\sqrt{2}} = \sin(\frac{2\pi}{6} t) = \sin(\frac{\pi}{3} t)$. Since $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$,we have $\frac{\pi}{4} = \frac{\pi}{3} t$. Solving for $t$: $t = \frac{3}{4} = 0.75 \ s$.

$A$ particle starts from the mean position and performs $S.H.M.$ with a period of $6 \ s$. At what time is its kinetic energy $50 \%$ of its total energy (in $s$)? $\left(\cos 45^{\circ} = \frac{1}{\sqrt{2}}\right)$

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