A block of mass $2\ \text{kg}$ 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 $2\ \text{m}$ and its spring constant is $200\ \text{N/m}$. The block is pushed such that the length of the spring becomes $1\ \text{m}$ and then released. At distance $x\ \text{m}$ ($x < 2$) from the wall, the speed of the block will be:
Answer: (B) $10[1 - (2 - x)^2]^{1/2}\ \text{m/s}$
Initial compression $1\ \text{m}$; at distance $x$ from the wall the compression is $(2 - x)$. Energy conservation:
$$\tfrac12k(1)^2 = \tfrac12mv^2 + \tfrac12k(2 - x)^2 \Rightarrow v^2 = \frac km\left[1 - (2 - x)^2\right] = 100\left[1 - (2 - x)^2\right]$$
$$v = 10\left[1 - (2 - x)^2\right]^{1/2}\ \text{m/s}$$
Solution by Sreeraj P, M.Sc Physics