$A_{R}=\sqrt{(\sqrt{3})^{2}+(1)^{2}}=2$

$\theta=\tan ^{-1}\left(\frac{\sqrt{3}}{1}\right)=\frac{\pi}{3}$

$\therefore y=2 \sin \left(\omega t+\frac{\pi}{3}\right)$

$\frac{d^{2} y}{d t^{2}}=a=-2 \omega^{2} \sin \left(\omega t+\frac{\pi}{3}\right)$

$a_{\max }=-2 \omega^{2}=g$

For which mass just breaks off the plank $\omega=\sqrt{g / 2}$

This will be happen for the first time when

$\omega t+\frac{\pi}{3}+\frac{\pi}{2}$ or $\omega t=\frac{\pi}{6}$

$\therefore t=\frac{\pi}{6 \omega}=\frac{\pi}{6} \sqrt{\frac{2}{g}}$