(a) Here, $a = 1$ cm =0.01 m; The mass will remain in contact with surface, if
$mg = m\omega^{2}a\ \ or\ \ \ \omega\sqrt{g/a}$
or $2\pi v = \sqrt{g/a}$ or $v = \frac{1}{2\pi}\sqrt{\frac{g}{a}}$
$= \frac{7}{2 \times 22}\sqrt{\frac{980}{1}} = 4.9\ Hz = 5\ Hz$