Vertical oscillations

$\large{y^{''} = -\omega^2 y = -\omega^2 [sin\omega t + \sqrt{3} cos\omega t]}$

Find max magnitude of $y^{''}$ and equate it to $g$

$\large{y^{''}_{max} = \omega^2 \sqrt{1 + 3} = 2 \omega^2 = g \\ \boxed{\omega_{min} = \frac{\sqrt{g}}{\sqrt{2}}}}$

The block can only leave the plank traveling downward. Set $y^{''}$ equal to $-g$ and find the associated time.

$\large{-g = -\frac{g}{2} [sin \, \omega t_{leave} + \sqrt{3} cos \, \omega t_{leave}] \\ sin\,\omega t_{leave} + \sqrt{3} cos\,\omega t_{leave} = 2 \\ \implies \omega t_{leave} = \frac{\pi}{6}} = \frac{\sqrt{g}}{\sqrt{2}} t_{leave} \\ \boxed{\large{t_{leave} = \frac{\sqrt{2} \pi}{6 \sqrt{g}}}}$