Question

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

• A. 0.1556 sec
• B. 0.1456 sec
• C. 0.1356 sec
• D. 0.1256 sec

Answer 1

Answer: (D)

As the platform is executing $SHM$, its time period will be minimum when it has the maximum acceleration. We know that in $SHM$, the maximum acceleration is given by
$a_{\max }={\omega }^{2}A$
Now if the body is not to be detached from the platform, $a_{\max }$ should be less than the acceleration due to gravity. In the limiting case,
$a_{\max }=g$
$\Rightarrow {\omega }^{2}A=g$
$\Rightarrow \frac{4{\pi }^{2}A}{T^2}=g$
$\Rightarrow T=2\pi \sqrt{\frac{A}{g}}=2\pi \times \sqrt{\frac{3.92\times 10^{-4}}{9.8}}$
$\therefore T_{\min }=0.1256sec$