For a particle executing simple harmonic motion, the kinetic energy K is given by K=K_0\cos ^{2}wt the maximum value of potential energy is

  • Option 1)

    K_0

  • Option 2)

    Zero

  • Option 3)

    \frac{K_0}{2}

  • Option 4)

    Not obtainable

 

Answers (1)

As we discussed in concept

Average value of kinetic energy -

Average of kinetic energy

=\frac{\int \left ( K.E. \right )\; dt}{ \int dt}
 

 

- wherein

= \frac{\int \frac{1}{2}m\omega ^{2}A^{2}\cos ^{2}\left ( \omega t+\phi \right )}{\int dt}

= \frac{1}{4}m\omega ^{2}A^{2}

 

 E_{K}\:=\:K_{o}cos^{2}wt

\therefore (E_{K})_{max}\:=\:K_{o}

(E_{p})_{max}=K_{o}                                                           \therefore (E_{tot})=K_{o}  


Option 1)

K_0

This option is correct.

Option 2)

Zero

This option is incorrect.

Option 3)

\frac{K_0}{2}

This option is incorrect.

Option 4)

Not obtainable

This option is incorrect.

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