Question

Vertical displacement of a plank with a body of mass ' $m$ ' on it is varying according to law $y=\sin \omega t+\sqrt{3} \cos \omega t$. The minimum value of $\omega$ for which the mass just breaks off the plank and the moment it occurs first after $t =0$ are given by: ( $y$ is positive vertically upwards)

• A. $\sqrt{\frac{ g }{2}}, \frac{\sqrt{2}}{6} \frac{\pi}{\sqrt{ g }}$
• B. $\frac{ g }{\sqrt{2}}, \frac{2}{3} \sqrt{\frac{\pi}{ g }}$
• C. $\sqrt{\frac{ g }{2}}, \frac{\pi}{3} \sqrt{\frac{2}{ g }}$
• D. $\sqrt{2 g }, \sqrt{\frac{2 \pi}{3 g }}$

Answer 1

Answer: (A)

$y =\sin \omega t +\sqrt{3} \sin (\omega t +\pi / 2)$
Amplitude, $A =\sqrt{1^{2}+(\sqrt{3})^{2}}=2$
$\tan \phi=\sqrt{3} \Rightarrow \phi=\frac{\pi}{3}$
$m \omega^{2} A = mg$
$\omega=\sqrt{\frac{ g }{ A }}=\sqrt{\frac{ g }{2}}$
$t =\frac{\pi / 6}{\omega}$