Question

A horizontal platform with an object placed on it is executing simple harmonic motion in the vertical direction. The amplitude of oscillation is $3.92\times {10}^{-3}m$. What should be the least period of these oscillations, so that the object is not detached from the platform?

• A. $0.145$ sec
• B. $0.1556$ sec
• C. $0.1256$ sec
• D. $0.1356$ sec

Answer 1

The correct option is C $0.1256$ sec

The possibility of the object to be detached from the platform is at the highest point when the platform starts going down. If it is less than $g$ then object will not get away from the platform. So, maximum acceleration of platform under $SHM$ is $g$.
From formula, maximum acceleration under $SHM={\omega }^{2}a$
${\omega }^{2}a=g\Rightarrow {\left(\frac{2\pi }{T}\right)}^2\times 3.92\times {10}^{-3}=1.0$
$T=2\pi \sqrt{\frac{3.92\times {10}^{-3}}{10}}=2\pi \sqrt{3.92\times {10}^{-4}}$$=0.1256$ secs