In the arrangement shown, the pulleys are small and light and the springs are ideal and ${\mathrm{K}}_1=25{\mathrm{\pi }}^2\mathrm{N}/\mathrm{m};{\mathrm{K}}_2=2{\mathrm{K}}_1;{\mathrm{K}}_3=3{\mathrm{K}}_1\text{ and }{\mathrm{K}}_4=4{\mathrm{K}}_1$ are the force constants of the springs.
Calculate the period of small oscillations of block of mass m =3 kg.

In static equilibrium of block , tension in the string is exactly equal to its weight. Let a vertically downward force F be applied on the block to pull it downward. Equilibrium is again restored when tension in the string is increases by the ame amount F. Hence total tension in the string becomes equal to (mg+F)
Strings are further elongated due to the extra tension F. Due to the extra tension F in string tension in each spring increases by 2F. Hence increase in elongation of springs is $2\mathrm{F}/{\mathrm{K}}_{1}2\mathrm{F}/{\mathrm{K}}_{2}2\mathrm{F}/{\mathrm{K}}_3\text{ and }2\mathrm{F}/{\mathrm{K}}_4$ respectively. According to geometry of the arrangement ,downward displacement of the block from its equilibrium position is
$\mathrm{y}=2\left(\frac{2\mathrm{F}}{{\mathrm{K}}_1}+\frac{2\mathrm{F}}{{\mathrm{K}}_2}+\frac{2\mathrm{F}}{{\mathrm{K}}_3}+\frac{2\mathrm{F}}{{\mathrm{K}}_4}\right)$   .....(i)
In the block is released now, is released now, it starts to accelerate upwards due to
extra tension f in the strings. It means restoring force on the block is equal to F. From
Eq(i)
$\mathrm{F}=\frac{\mathrm{y}}{4\left(\frac{1}{{\mathrm{K}}_1}+\frac{1}{{\mathrm{K}}_2}+\frac{1}{{\mathrm{K}}_3}+\frac{1}{{\mathrm{K}}_4}\right)}$
Restoring acceleration of block
$\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{y}}{4\mathrm{m}\left(\frac{1}{{\mathrm{K}}_1}+\frac{1}{{\mathrm{K}}_2}+\frac{1}{{\mathrm{K}}_3}+\frac{1}{{\mathrm{K}}_4}\right)}$
Since acceleration of block is restoring and is directly proportional to displacement ,the block performs SHM.
Its period , $\mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\text{ displacement }}{\text{ acceleration }}}$
$\begin{array}{r}\mathrm{T}=2\mathrm{\pi }\sqrt{4\mathrm{\pi }\left(\frac{1}{{\mathrm{K}}_1}+\frac{1}{{\mathrm{K}}_2}+\frac{1}{{\mathrm{K}}_3}+\frac{1}{{\mathrm{K}}_4}\right)}\\ \mathrm{T}=4\mathrm{\pi }\sqrt{\mathrm{m}\left(\frac{1}{{\mathrm{K}}_1}+\frac{1}{{\mathrm{K}}_2}+\frac{1}{{\mathrm{K}}_3}+\frac{1}{{\mathrm{K}}_4}\right)}\end{array}$
After substituting the values, we get T = 2 s.