A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is 200 N/m. The block is pushed such that the length of the spring becomes 1 m and then released. At distance x m (x < 2) from the wall, the speed of the block will be :

Let the natural (unstretched) length of the spring be $$L_0 = 2\,\text{m}$$.
The free end of the spring is fixed to the wall and the block of mass $$m = 2\,\text{kg}$$ is attached to the other end.

When the spring is compressed to a length $$1\,\text{m}$$ and released, the compression produced is
$$y_{\text{max}} = L_0 - 1 = 2 - 1 = 1\,\text{m}$$.
This maximum compression is the amplitude $$A$$ of the ensuing simple harmonic motion (SHM):
$$A = 1\,\text{m}$$.

The force constant (spring constant) is $$k = 200\,\text{N\,m}^{-1}$$, so the angular frequency of the SHM is
$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{2}} = \sqrt{100} = 10\,\text{rad\,s}^{-1}$$.

At any instant, let the actual length of the spring (i.e., the distance of the block from the wall) be $$x$$ metres.
Because $$x \lt 2\,\text{m}$$, the spring is compressed.
Define the instantaneous compression (displacement from the natural length) as
$$y = L_0 - x = 2 - x.$$

For SHM with amplitude $$A$$, angular frequency $$\omega$$ and displacement $$y$$ from equilibrium (natural length here), the speed $$v$$ is related by the standard formula
$$v = \omega \sqrt{A^{2} - y^{2}}\;.$$

Substituting $$\omega = 10\,\text{rad\,s}^{-1}$$, $$A = 1\,\text{m}$$ and $$y = 2 - x$$:
$$v = 10 \sqrt{1^{2} - (2 - x)^{2}} = 10\,[\,1 - (2 - x)^{2}\,]^{1/2}\,\text{m\,s}^{-1}.$$

Hence the speed of the block when it is at a distance $$x$$ from the wall is given by
Option B: $$\displaystyle 10[1 - (2 - x)^{2}]^{1/2}\,\text{m\,s}^{-1}\;.$$