Explanation

(b) Here $a = 12\ cms^{- 2},\ x = 3\ cm$
In SHM, acceleration $a = - \omega^{2}x$
$\therefore\ $Magnitude of acceleration $a = \omega^{2}x$
(discarding off –ve sign)
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \omega^{2} = \frac{a}{x}$
or $\omega = \sqrt{\frac{a}{x}}$
or $\frac{2\pi}{T} = \omega = \sqrt{\frac{a}{x}}$
or $T = 2\pi\sqrt{\frac{a}{x}}$
$= 2\pi\sqrt{\frac{3}{12}}$
$= \pi s = 3.14$ s