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  • Help me solve this A particle of mass 10 g is describing SHM along a straight line with period of 2 s and amplitude of 10 cm. Its kinetic energy when it is at 5 cm from its equilibrium position is

A particle of mass 10 g is describing SHM along a straight line with period of 2 s and amplitude of 10 cm. Its kinetic energy when it is at 5 cm from its equilibrium position is

  • Option 1)

    37.5\pi^{2} \;ergs

  • Option 2)

    3.75\pi^{2} \;ergs

  • Option 3)

    375\pi^{2} \;ergs

  • Option 4)

    0.375\pi^{2} \;ergs

 

Answers (1)

best_answer

Kinetic energy

 \\*k = \frac{1}{2}m\omega^{2}(a^{2}-y^{2}) \\* = \frac{1}{2}\times 10\times (\frac{2\pi}{2})^{2}[10^{2} - 5^{2}] = 375\pi^{2}\;ergs

 

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}

 

 

 


Option 1)

37.5\pi^{2} \;ergs

This is incorrect.

Option 2)

3.75\pi^{2} \;ergs

This is incorrect.

Option 3)

375\pi^{2} \;ergs

This is correct.

Option 4)

0.375\pi^{2} \;ergs

This is incorrect.

Posted by

prateek

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