Question

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

• A. $0.1256\text{sec}$
• B. $0.1356\text{sec}$
• C. $0.1456\text{sec}$
• D. $0.1556\text{sec}$

Answer 1

The correct option is A $0.1256\text{sec}$


From FBD of object,
$mg-R=ma$
where, $m=$ mass of the object,
$R=$ Reaction force,
$a=$ acceleration of the body,
$\omega =$ angular frequency of SHM.
For critical condition of object just losing contact with platform surface, $R=0$.
So, $mg=ma$
$\Rightarrow mg=mA{\omega }^2$
where $A=$ amplitude of SHM [since $A{\omega }^2$ is the max. acceleration]
$\Rightarrow \omega =\sqrt{g/A}=\sqrt{\frac{9.8}{3.92\times {10}^{-3}}}=50\text{rad/s}$
$\Rightarrow T=\frac{2\pi }{\omega }=\frac{2\pi }{50}=0.1256\text{sec}$
Hence, option $\left(a\right)$ is the correct answer.