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  • Solve this problem A particle is executing simple harmonic motion with a time period T. At time t=0, it is at its position of equilibrium. The kinetic energy - time graph of the particle will look like :

A particle is executing simple harmonic motion with a time period T.  At time t=0, it is at its position of equilibrium.  The kinetic energy - time graph of the particle will look like :

 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 

Answers (1)

As we learnt in

Total energy in S.H.M. -

Total Energy = Kinetic + Potential Energy

- wherein

Total Energy =\frac{1}{2}K\left ( A^{2}-x^{2} \right )+\frac{1}{2}kx^{2}= \frac{1}{2}kA^{2}

Hence total energy in S.H.M. is constant

 

 At mean position  t = 0, \omega t=0,  y = 0

v=v_{max}=a\omega

\therefore\ \;K.E.=K.E_{max}=\frac{1}{2}m\omega^{2}a^{2}

At extreme position t=\frac{T}{4},\ \; \omega=\frac{\pi}{2},\ \; y=A

v_{min}=0\ \therefore\ \; K.E.=K.E_{min}=0

K.E. in S.H.M. = K.E.=\frac{1}{2}m\omega^{2}(a^{2}-y^{2})

   y=a\ sin\omega t

K.E.=\frac{1}{2}m\omega^{2}(a^{2}-a^{2}sin^{2}\omega t)=\frac{1}{2}m\omega^{2}a^{2}(1-sin^{2}\omega t)

\therefore\ \;K.E.=\frac{1}{2}m\omega^{2}a^{2}cos^{2}\omega t

Correct option is 4.

 


Option 1)

This is an incorrect option.

Option 2)

This is an incorrect option.

Option 3)

This is an incorrect option.

Option 4)

This is the correct option.

Posted by

Sabhrant Ambastha

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