The potential energy of a simple harmonic oscillator of mass $2\, kg$ in its mean position is $5\, J.$ If its total energy is $9\,J$ and its amplitude is $0.01\, m,$ its time period would be
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kinetic energy at mean position $=$ total $energy-potential$ energy at mean position$=$
$9 J-5 J=4 J$
kinetic energy at mean position $=\frac{1}{2} m v_{\max }^{2}$
$\Rightarrow \frac{1}{2} m v_{\max }^{2}=4 J$
$\Rightarrow v_{\max }=\sqrt{\frac{4 \times 2}{2}} \mathrm{m} / \mathrm{s}=2 \mathrm{m} / \mathrm{s}$
$\Rightarrow A \omega=A \frac{2 \pi}{T}$
$\Rightarrow A \frac{2 \pi}{T}=2 m / s$
$\Rightarrow T=A \pi=0.01 \pi=\frac{\pi}{100} s$
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