The time periods of systems depicted below under identical conditions are respectively

i)$2\pi \sqrt{\frac{\text{m}}{\text{k}}}$ ;       ii) $2\pi \sqrt{\frac{\text{4m}}{\text{k}}}$ ;         iii) $2\pi \sqrt{\frac{\text{m}}{\text{4k}}}$

Fig (a) : We know that

$\text{T}=\text{2}\pi \sqrt{\frac{\text{m}}{\text{k}}}$

Fig (b) : $\text{K}{x}_0=\text{mg;}\text{ma}=-\text{K}{x}_0\Rightarrow \text{a}=\frac{\text{K}}{\text{m}}{x}_0$

But$\text{a}=-{\omega }^2{x}_0$

$\omega =\sqrt{\frac{\text{K}}{\text{m}}}\Rightarrow \text{T}=2\pi \sqrt{\frac{\text{m}}{\text{k}}}$

So, the extra force in spring will be$\text{k}x\text{/2}$ , as the pulley is massless, this

force is equal to 2T.

$2\text{T}=\frac{\text{k}x}{2}\Rightarrow \text{T}=\frac{\text{k}x}{4}$

Restoring force$\text{F}=-\frac{\text{k}x}{4};\text{a}=\frac{-\text{k}}{4\text{m}}x$

$\text{T}=2\pi \sqrt{\frac{x}{\text{a}}}\Rightarrow \text{T}=2\pi \sqrt{\frac{\text{4m}}{\text{k}}}$

$\text{F}=-4\text{K}x;\text{a}=-\frac{\text{4k}}{\text{m}}x$

$\text{T}=2\pi \sqrt{\frac{x}{\text{a}}}\Rightarrow \text{T}=2\pi \sqrt{\frac{\text{m}}{\text{4k}}}\left(i,i,ii,iii\right)$