A block of mass 2kg is attached with two identical springs of...

When the block is displaced by a distance $x$, force acting on block due to both springs will be in the same direction. Therefore,

$F = - 2kx ⇒ a = - \dfrac{2k}{m}x,$

Comparing it with$a = - {\omega}^{2}x$ , we get

$\omega = \sqrt{\dfrac{2k}{m}} = \sqrt{\dfrac{2 \times 20}{2}}$

$= \sqrt{20}rad{s}^{- 1}$

Now, $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{\sqrt{20}} = \dfrac{\pi}{\sqrt{5}}$

Hence, required value of $X = 5$.