A

Step 1: Calculate the initial potential energy stored in the spring when compressed to one-fourth of its natural length: \(PE = \frac{1}{2} k (L_0 - L)^2\), where \(L = \frac{L_0}{4}\), yielding \(PE = \frac{1}{2} k \left(L_0 - \frac{L_0}{4}\right)^2 = \frac{1}{2} k \left(\frac{3L_0}{4}\right)^2 = \frac{9}{32} k L_0^2\).
Step 2: Use conservation of energy to relate potential energy to kinetic energy when the block is at distance \(x\): \(\frac{9}{32} k L_0^2 = \frac{1}{2} mv^2 + \frac{1}{2} k (x - L_0)^2\). Rearranging gives \(v^2 = \frac{9 k L_0^2 - k (x - L_0)^2}{16 m}\). Thus, the velocity as a function of \(x\) is given by \(v(x) = \sqrt{\frac{9 k L_0^2 - k (x - L_0)^2}{16 m}}\).
Step 3: The maximum velocity occurs when \(x = L_0\): \(v_{max} = \sqrt{\frac{9 k L_0^2}{16 m}}\). Therefore, \(v_{max} = \frac{3}{4} \sqrt{\frac{k}{m}} L_0\).
Thus, the velocity as a function of distance \(x\) from the wall is given and the maximum velocity of the block is also derived.