College Physics (4th Edition)

$a = 7.9~m/s^2$
We can find $\omega$: $T = \frac{2\pi}{\omega}$ $\omega = \frac{2\pi}{T}$ $\omega = \frac{2\pi}{0.50~s}$ $\omega = 4\pi~rad/s$ At the point of maximum extension of the spring, the magnitude of acceleration is at its maximum. We can find the magnitude of the acceleration: $a = A~\omega^2$ $a = (0.050~m)~(4\pi~rad/s)^2$ $a = 7.9~m/s^2$