In the arrangement shown in figure, pulleys ate small and light and springs are ideal .$K_1=25{\pi }^2,K_2=2K_1,$ $K_3=3K_1\text{ and }{K}_4=4K_1$ are force constants of the springs. The period of small vertical oscillations of block of mass, m=3kg is _________

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 string is increased by the same amount F. Hence, total tension in string becomes equal to (mg + F).

Springs are further elongated due to the extra tension F. Due to this extra tension F in strings, tension in each spring increase by 2F. Hence increase in elongation of springs is $\frac{2F}{K_1},\frac{2F}{K_2},\frac{2F}{K_3}\text{ and }\frac{2F}{K_4}\text{ respectively. }$

According to geometry of the arrangement, downward displacement of the block from its equilibrium position is, 

$y=2\left(\frac{2F}{K_1}+\frac{2F}{K_2}+\frac{2F}{K_3}+\frac{2F}{K_4}\right)$………(i)

If the block is released now, it starts to accelerate upwards due to extra tension F in strings. It means restoring force on the block is equal to F.

From equation (i), $F=\frac{y}{4\left(\frac{1}{K_1}+\frac{1}{K_2}+\frac{1}{K_3}+\frac{1}{K_4}\right)}$

Restoring acceleration of block

$\frac{F}{m}=\frac{y}{4m\left(\frac{1}{K_1}+\frac{1}{K_2}+\frac{1}{K_3}+\frac{1}{K_4}\right)}$

Since, acceleration of block is restoring and is directly proportional to displacement y, therefore, the block performs SHM.

$\text{ Its period, }T=2\pi \sqrt{\frac{\text{ displacement }}{\text{ acceleration }}}$

$\begin{array}{l}T=2\pi \sqrt{4m\left(\frac{1}{K_1}+\frac{1}{K_2}+\frac{1}{K_3}+\frac{1}{K_4}\right)}\\ T=4\pi \sqrt{m\left(\frac{1}{K_1}+\frac{1}{K_2}+\frac{1}{K_3}+\frac{1}{K_4}\right)}\end{array}$

After substituting the values, we get T = 2 s.