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(B) The given equation for $S.H.M.$ is $y = A \sin(\omega t + \phi)$,where $A = 2 \, cm$ and $\omega = \frac{\pi}{2} \, rad/s$.The velocity $v$ is giv

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$\frac{\pi^2}{2} \, cm/s^2$

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(B) The given equation for $S.H.M.$ is $y = A \sin(\omega t + \phi)$,where $A = 2 \, cm$ and $\omega = \frac{\pi}{2} \, rad/s$.The velocity $v$ is given by $\frac{dy}{dt} = A\omega \cos(\omega t + \phi)$.The acceleration $a$ is given by $\frac{d^2y}{dt^2} = -A\omega^2 \sin(\omega t + \phi)$.The maximum acceleration $a_{max}$ is given by $|A\omega^2|$.Substituting the values: $a_{max} = 2 	imes \left( \frac{\pi}{2} ight)^2 = 2 	imes \frac{\pi^2}{4} = \frac{\pi^2}{2} \, cm/s^2$.

For a particle performing $S.H.M.$ given by the equation $y = 2 \sin \left( \frac{\pi t}{2} + \phi \right) \, (cm)$,what is the maximum acceleration of the particle?

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