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(A) The acceleration of a particle in simple harmonic motion is given by $f = -\omega^2 x$. Given the acceleration-time graph,at $t = 0$,$f$ is negati

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(A) The acceleration of a particle in simple harmonic motion is given by $f = -\omega^2 x$. Given the acceleration-time graph,at $t = 0$,$f$ is negative and at its maximum magnitude. This corresponds to the particle being at the positive extreme position $(x = A)$. The equation for displacement is $x = A \cos(\omega t)$. The velocity is $v = \frac{dx}{dt} = -A\omega \sin(\omega t)$. At $t = 0$,$v = 0$. At $t = T/4$,$v = -A\omega$ (minimum). At $t = T/2$,$v = 0$. At $t = 3T/4$,$v = A\omega$ (maximum). Comparing this with the given options,the graph that starts at $0$ at $t=0$,goes to a negative minimum at $t=T/4$,passes through $0$ at $t=T/2$,and reaches a positive maximum at $t=3T/4$ is Graph $A$.

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