The spring's mass isn't negligible?

Since the spring has mass and a uniform density, integrating each differential kinetic energy $dT$ is the best approach.

$\displaystyle dT = \frac{1}{2}u^2 dm$

$\displaystyle T_s = \int_{m} \frac{1}{2}u^2 dm$

Where $\displaystyle dm = \frac{m}{L} dx$ , and $\displaystyle u = \frac{x}{L} v$

$\displaystyle T_s = \frac{1}{2} \frac{m}{L} \int_{0}^{L} \left(\frac{vx}{L}\right)^2 dx$

$\displaystyle T_s = \frac{1}{2} \frac{m}{L^3} v^2 \frac{L^3}{3}$

This leaves the kinetic energy of the spring as $\displaystyle \boxed{\frac{mv^2}{6}}$ . The rest is pretty straightforward, adding the kinetic energy of the mass to the expression.

The resultant expression is $\displaystyle T(v) = \frac{1}{2} \left(M + \frac{1}{3}m \right)v^2$

( WOW ) I did a solution, it turned out wrong, then I had 1 chance and I shook it and it turned out to be correct.