MCQ
A horizontal platform with an object placed on it is executing $S.H.M$. in the vertical direction. The amplitude of oscillation is $3.92 \times {10^{ - 3}}m$. What must be the least period of these oscillations, so that the object is not detached from the platform
  • $0.1256\, sec$
  • B
    $0.1356\, sec$
  • C
    $0.1456\, sec$
  • D
    $0.1556\, sec$

Answer

Correct option: A.
$0.1256\, sec$
a
(a) 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\,sec$

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