The graph shows the variation of displacement | vedclass

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(D) From the graph,at time $t = T/2$,the displacement $y$ is at its negative extreme position $(-A)$.At the extreme position,the velocity of the parti

Vedclass · Accepted Answer

The $P$.$E$. is equal to total energy at time $T/2$

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(D) From the graph,at time $t = T/2$,the displacement $y$ is at its negative extreme position $(-A)$.At the extreme position,the velocity of the particle is zero $(v = 0)$.The total energy $(E)$ of a particle in $S$.$H$.$M$. is the sum of kinetic energy $(K.E.)$ and potential energy $(P.E.)$,given by $E = K.E. + P.E.$Since $K.E. = \frac{1}{2}mv^2$,at $v = 0$,$K.E. = 0$.Therefore,at the extreme position,$E = P.E.$Thus,the potential energy is equal to the total energy at time $T/2$.

The graph shows the variation of displacement of a particle executing $S$.$H$.$M$. with time. We infer from this graph that

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