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A particle starts from mean position and performs S.H.M. with period 6 second. At what time its kinetic energy is $50 \%$ of total energy? $\left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)$
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The correct answer is:
$0.75$ second
$T=6 \mathrm{~sec}$
$\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\frac{1}{2}\left(\frac{1}{2} m \omega^{2} A^{2}\right.)$
$A^{2}-x^{2}=\frac{A^{2}}{2}$
$\therefore x^{2}=\frac{A^{2}}{2} \Rightarrow x=\frac{A}{\sqrt{2}}$
$\frac{A}{\sqrt{2}}=A \sin \omega t=A \sin \frac{2 \pi}{T} t$
$\frac{1}{\sqrt{2}}=\sin \frac{2 \pi}{6} t=\sin \frac{\pi}{3} t$
$\sin \frac{\pi}{4} t=\sin \frac{\pi}{3} t$
$\therefore t=\frac{3}{4}=0.75 \mathrm{sec}$
$\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\frac{1}{2}\left(\frac{1}{2} m \omega^{2} A^{2}\right.)$
$A^{2}-x^{2}=\frac{A^{2}}{2}$
$\therefore x^{2}=\frac{A^{2}}{2} \Rightarrow x=\frac{A}{\sqrt{2}}$
$\frac{A}{\sqrt{2}}=A \sin \omega t=A \sin \frac{2 \pi}{T} t$
$\frac{1}{\sqrt{2}}=\sin \frac{2 \pi}{6} t=\sin \frac{\pi}{3} t$
$\sin \frac{\pi}{4} t=\sin \frac{\pi}{3} t$
$\therefore t=\frac{3}{4}=0.75 \mathrm{sec}$
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