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  • Solve it, 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

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) s/12

     

  • Option 2) s/6

     

  • Option 3) s/4

     

  • Option 4) s/3

     

 

Answers (1)

best_answer

As we learnt in

Kinetic energy in S.H.M. -

K.E.= \frac{1}{2}mu^{2}\\ \: \: \: = \frac{1}{2}m\left ( A^{2} -x^{2}\right )\omega ^{2}

 

- wherein

K.E.= \frac{1}{2}K\left ( A^{2}-x^{2} \right ) \\K= m\omega ^{2}

 

 

 

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)

\frac{1}{12}s

This is an incorrect option.

Option 2)

\frac{1}{6}s

This is the correct option.

Option 3)

\frac{1}{4}s

This is an incorrect option.

Option 4)

\frac{1}{3}s

This is an incorrect option.

Posted by

divya.saini

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