1. Identify equilibrium and initial condition

The spring has natural length $L_0 = 2\ \text{m}$ and spring constant $k = 200\ \text{N/m}.$

The block of mass $m = 2\ \text{kg}$ is released when the spring length is $1\,$m.

So the compression from natural length is $A = 2-1 = 1\ \text{m}.$

Since the table is frictionless and the block is released from rest at maximum compression, the motion is SHM about the equilibrium position at spring length $2\,$m with amplitude $A=1\ \text{m}.$

2. Displacement at a general position

When the block is at distance $x$ from the wall, the spring length is $x$.

Hence displacement from equilibrium is $y = x-2.$

Its magnitude is $|y| = 2-x$ because $x<2$.

3. Use energy conservation / SHM speed formula

For SHM, $v = \omega\sqrt{A^2-y^2}$ where $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{2}} = \sqrt{100} = 10\ \text{rad/s}.$

Now, $A=1, \qquad y=x-2.$

So $v = 10\sqrt{1-(x-2)^2}.$

Since $(x-2)^2=(2-x)^2$, $v = 10\sqrt{1-(2-x)^2}.$

Thus the speed is $\boxed{10\left[1-(2-x)^2\right]^{1/2}\ \text{m/s}}.$

4. Check options

• A: $10\left[1-(2-x)^2\right]^{1/2}$ ✅
• B: power $3/2$ ❌
• C: power $1$ ❌
• D: power $2$ ❌

So the correct option is A.