Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy by 75% of the total energy Option 1) Option 2) Option 3) Option 4)

As we learnt in

Kinetic energy in S.H.M. -

- wherein

During simple harmonic motion, Kinetic energy

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

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

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

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

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

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

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

Correct option is 2.

Option 1)

This is an incorrect option.

Option 2)

This is the correct option.

Option 3)

This is an incorrect option.

Option 4)

This is an incorrect option.

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