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. $\sqrt{2g},\sqrt{\frac{2\pi }{3g}}$

Answer 1

Answer: (A)

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