Vertical displacement of a plank with a body of mass $m$ on it is varying according to the law $y= \sin{\omega t} + \sqrt{3}\cos{\omega t}$ . The minimum value of $\omega$ such that the mass breaks off the plank and the moment it occur first after $t=0$ is given by (y is positive vertically upwards)

$\omega = \dfrac{\sqrt{ag}}{\sqrt{b}}$ and $t=$ $\dfrac{\sqrt{z}\pi}{6\sqrt{g}}$ respectevly. Find the value of $a+b+z$

HERE a and b are positive coprime integers which are square free

The answer is 5.

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

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.

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