The acceleration A and time t of a body in S. | vedclass

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(A) In $S.H.M.$,the acceleration $A$ is given by $A = -\omega^2 x$. The given graph shows $A$ starting from a negative maximum value at $t = 0$. This

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(A) In $S.H.M.$,the acceleration $A$ is given by $A = -\omega^2 x$. The given graph shows $A$ starting from a negative maximum value at $t = 0$. This implies the displacement $x$ is at a positive maximum (extreme position) at $t = 0$.At the extreme position,the velocity $v$ of the particle is zero,and therefore the kinetic energy $(K.E. = \frac{1}{2}mv^2)$ is zero.As the particle moves towards the mean position,its speed increases,reaching a maximum at the mean position,where acceleration is zero. Thus,$K.E.$ becomes maximum when acceleration is zero.Since the acceleration completes one full cycle in time $T$,the kinetic energy,which depends on $v^2$,completes two cycles in the same time $T$. The graph starting from zero at $t = 0$,reaching a maximum,and returning to zero at $t = T/2$ corresponds to the behavior of $K.E.$ in $S.H.M.$ This is correctly represented by the graph in option $A$.

The acceleration $A$ and time $t$ of a body in $S.H.M.$ is given by the curve shown below. Then the corresponding graph between kinetic energy $(K.E.)$ and time $t$ is correctly represented by:

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