c
(c)$n = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}} $
$\Rightarrow n \propto \frac{1}{{\sqrt m }} $
$\Rightarrow \frac{{{n_1}}}{{{n_2}}} = \sqrt {\frac{{{m_2}}}{{{m_1}}}} $
$ \Rightarrow \frac{n}{{{n_2}}} = \sqrt {\frac{{4m}}{m}}$
$4\Rightarrow {n_2} = \frac{n}{2}$