Spring-Mass System

$F = kx \Rightarrow 10 = k(5 \times 10^{-2}) \Rightarrow k = 200$ N/m $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2}{200}} = \frac{2\pi}{10} = 0.628$ s

Time period of a spring–mass system: $T = 2\pi\sqrt{\frac{M}{k}}$, so $T \propto \sqrt{M}$
When another mass M is added, the total mass becomes 2M.
$\frac{T_2}{T_1} = \sqrt{\frac{M + M}{M}} = \sqrt{2}$
$T_2 = \sqrt{2}T$

The mass can get detached from the pan only at the upper extreme position, where the acceleration is directed downward and has its maximum value $\omega^2 a$.
At that point, taking downward as positive: $mg - R = m\omega^2 a$, where $R$ is the normal reaction of the pan.
$R = mg - m\omega^2 a$
The mass just loses contact when $R = 0$, i.e. $g = \omega^2 a$
$\Rightarrow a = \dfrac{g}{\omega^2} = \dfrac{mg}{K}$, as $\omega^2 = \dfrac{K}{m}$
$\Rightarrow a = \dfrac{2 \times 10}{200}$
$\Rightarrow a = \dfrac{1}{10}$ m $= 10$ cm
So the minimum amplitude is 10.0 cm.

For a spring, the spring constant is inversely proportional to its length: $kl$ = constant.
Each quarter has length $\dfrac{l}{4}$, so its spring constant is $k' = 4k$.
Time period: $T = 2\pi\sqrt{\dfrac{m}{k}}$
$T' = 2\pi\sqrt{\dfrac{m}{4k}} = \dfrac{1}{2} \times 2\pi\sqrt{\dfrac{m}{k}} = \dfrac{T}{2}$
So the new time period is $\dfrac{T}{2}$.