starting from the origin a body oscillates simple harmonically with the period of 2 seconds after what time will kinetic energy be 75% of Total energy

Answers (1)

During simple harmonic motion, Kinetic energy

== \frac{1}{2}m\nu ^{2}= \frac{1}{2}m\left ( a\: \omega \cos \omega t \right )^{2}

Total \: energy \: E= \frac{1}{2}ma^{2}\omega ^{2}

\because \left ( Kinetic\: energy \right )= \frac{75}{100}(E)

or\: \: \: \frac{1}{2}ma^{2}\omega ^{2}\cos ^{2}\omega t= \frac{75}{100}\times \frac{1}{2}ma^{2}\omega ^{2}

or\: \: \cos ^{2}\omega t= \frac{3}{4}\Rightarrow \cos \omega t= \frac{\sqrt{3}}{2}= \cos \frac{\pi }{6}

\therefore \omega t= \frac{\pi }{6}

or\: \: \: t= \frac{\pi }{6\omega }= \frac{\pi }{6\left ( 2\pi/T \right ) }= \frac{2\pi }{6\times 2\pi }= \frac{1}{6}s

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