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 from the plank and the moment it occurs is first after \(t=0\), are given by:

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

Hint:

Separation occurs when normal reaction becomes zero: \[ a_{\text{down}}=g \] Find maximum downward acceleration.

Answer 1

The Correct Option is A

Step 1: Acceleration of plank: \[ a=-\omega^2y \]
Step 2: Maximum downward acceleration occurs at maximum \(y\): \[ y_{\max}=\sqrt{1+3}=2 \]
Step 3: Condition for separation: \[ \omega^2 y_{\max}=g \Rightarrow \omega=\sqrt{\frac{g}{2}} \]
Step 4: Phase angle: \[ \tan\phi=\sqrt{3}\Rightarrow \phi=\frac{\pi}{3} \]
Step 5: Time: \[ t=\frac{\pi/3}{\omega}=\frac{\sqrt{2}\pi}{6} \]