Problem 12.41

12.41. a)
For each segment of the spring,
$dK = \frac{1}{2} (dm)(v_{x})^{2}$
And the velocity at a position x away from the wall is given by
$v_{x} = \frac{x}{\ell} v$
where v is the velocity of the block. Now, we need an expression for the mass of the small segment of the spring dm
$dm = \frac{m}{\ell}dx$
Therefore the kinetic energy of the small section of the spring is given by
$dK = \frac{1}{2} \frac{x^{2}v^{2}m}{\ell^{3}} dx$
and the total kinetic energy is given by integrating over the length of the spring from 0 to $\ell$ and adding the kinetic energy of the block.
$K = \frac{1}{2} M v^{2} + \int_{0}^{\ell}\left(\frac{x^{2}v^{2}m}{\ell^{3}}\right) dx$
$K = \frac{1}{2} (M + \frac{m}{3}) v^{2}$

b)
So, we can consider the effective mass to be
$m_{eff} = (M + \frac{m}{3})$
and $\omega_{eff} = \frac{k}{m_{eff}}$
The period is given by
$T = 2\pi \sqrt{\frac{M + \frac{m}{3}}{k}}$