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 \(mg = mA\)

\( \Rightarrow mg = ma{\omega ^2}\)

\( \Rightarrow \omega = \sqrt {g/a} = \sqrt {\frac{{9.8}}{{3.92 \times {{10}^{ - 3}}}}} = 50\) .

\( \Rightarrow T = \frac{{2\pi }}{\omega } = \frac{{2\pi }}{{50}} = 0.1256\,sec\)