Question

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

• A. $\sqrt{\frac{g}{2}},\sqrt{\frac{2}{g}}\frac{\pi }{6}$
• 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)

The acceleration deu to SHM is given by $a=d^{2}y/d{t}^2=-{\omega }^2\times sin\left(\omega t\right)+cos\left(\sqrt{3}\omega t\right)$$=-2{\omega }^{2}sin(\pi /3+\omega t)$

When Amplitude of $a$ is equal to the acceleration due to gravity, the mass just breaks off the plank i.e., $-2{\omega }^2=-g\Rightarrow \omega =\sqrt{g/2}$

Time at which mass breaks off the plank is given by, $=-2{\omega }^{2}sin(\pi /3+\omega t)=-g\Rightarrow \omega t=\pi /6\Rightarrow t=\pi /6\omega =\sqrt{2}\pi /6\sqrt{g}$