In a simple harmonic oscillator, at the mean position

  • Option 1)

    kinetic energy is minimum, potential energy is maximum

  • Option 2)

    both kinetic and potential energies are maximum

  • Option 3)

    kinetic energy is maximum, potential energy is minimum

  • Option 4)

    both kinetic and potential energies are minimum.

 

Answers (1)

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}

 

 

Potential energy in S.H.M. -

P.E.= \frac{1}{2}Kx^{2}

 

- wherein

Where K= m\omega ^{2}

 

 

Hence, K.E. maximum and P.E. is minimum at mean position.

Correct option is 3.

 

 


Option 1)

kinetic energy is minimum, potential energy is maximum

This is an incorrect option.

Option 2)

both kinetic and potential energies are maximum

This is an incorrect option.

Option 3)

kinetic energy is maximum, potential energy is minimum

This is the correct option.

Option 4)

both kinetic and potential energies are minimum.

This is an incorrect option.

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