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 :

Answers (1)

As we learned,

The 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

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