Explanation

(c) Velocity $v = \omega\sqrt{A^{2} - x^{2}}$
and acceleration = $\omega^{2}x$
Given, $\omega\sqrt{A^{2} - x^{2}} = \omega^{2}x\ \ or\ \ \sqrt{A^{2} - x^{2}} = \omega x$
Given, $T = \ \frac{2\pi}{\sqrt{3}}$
and $\omega = \frac{2\pi}{T} = \sqrt{3}$
substituting the value of $\omega$ in Eq. (i), we get
$\sqrt{A^{2} - x^{2}} = \sqrt{3}x$
$\Rightarrow$ $A = 2x$
As amplitude= $\frac{path\ length}{2} \times 2\ cm$
$\Rightarrow$ $x = 1$ cm