A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is $3.92 \times 10^{-3} \mathrm{m}$ . What must be the least period of these oscillations, so that the object is not detached from the platform
Text Solution
Verified by ExpertsA
By drawing free body diagram of object during the downward motion at extreme position, for equilibrium of mass
$mg - R = mA$ (A = Acceleration)
For critical condition R = 0
so mg = mA $\Rightarrow mg = ma \omega^{2}$
$\Rightarrow \omega = \sqrt{g/a} = \sqrt{\frac{9.8}{3.92 \times 10^{-3}}} = 50$
$\Rightarrow T = \frac{2 \pi}{\omega} = \frac{2 \pi}{50} = 0.1256 \text{ sec}$
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