Springs Off The Wall (Fixed)

Here's a diff eq solution. A much more elementary numerical solution also yields the same answer.

Newton's 2nd Law Equations:

$k(x_2 - x_1 - l_0) = m_1 \ddot{x_1} \\ -k(x_2 - x_1 - l_0) = m_2 \ddot{x_2}$

Re-arranging:

$\frac{k}{m_1}(x_2 - x_1 - l_0) = \ddot{x_1} \\ -\frac{k}{m_2}(x_2 - x_1 - l_0) = \ddot{x_2} \\ \ddot{x_2} - \ddot{x_1} = -(\frac{k}{m_1} + \frac{k}{m_2})(x_2 - x_1) + l_0 (\frac{k}{m_1} + \frac{k}{m_2}) \\ \alpha = (\frac{k}{m_1} + \frac{k}{m_2})$

Simple Harmonic Motion - General Form:

$x_2 - x_1 = A cos(\sqrt{\alpha} t) + B sin(\sqrt{\alpha} t) + C \\ \dot{x_2} - \dot{x_1} = -A \sqrt{\alpha} sin(\sqrt{\alpha} t) + B \sqrt{\alpha} cos(\sqrt{\alpha} t) \\ \ddot{x_2} - \ddot{x_1} = -A \alpha cos(\sqrt{\alpha} t) - B \alpha sin(\sqrt{\alpha} t)$

Initial Velocity Conditions:
Call $t = 0$ the time at which the spring first reaches its equilibrium length.
Call $x_{2_0}$ the position of the second block before the string is cut.

$\frac{1}{2} k (l_0 - x_{2_0})^2 = \frac{1}{2} m_2 \dot{x_2}^2 \\ \dot{x_2} = \sqrt{\frac{k}{m_2}} (l_0 - x_{2_0}) \hspace{1cm} \text{@ t= 0} \\ \implies \dot{x_2} - \dot{x_1} = \sqrt{\frac{k}{m_2}} (l_0 - x_{2_0}) \hspace{1cm} \text{@ t= 0}$

Initial Position and Acceleration:

$x_2 - x_1 = l_0 \hspace{1cm} \text{@ t = 0} \\ \ddot{x_2} - \ddot{x_1} = 0 \hspace{1cm} \text{@ t = 0}$

Plugging Initial Conditions into General Form:

$l_0 = A + C \\ \sqrt{\frac{k}{m_2}} (l_0 - x_{2_0}) = B \sqrt{\alpha} \\ 0 = -A \alpha$

Solving for Parameters:

$A = 0 \\ B = \sqrt{\frac{k}{\alpha m_2}} (l_0 - x_{2_0}) \\ C = l_0$

Final Expression for the Stretch:

$x_2 - x_1 = \sqrt{\frac{k}{\alpha m_2}} (l_0 - x_{2_0}) \, sin(\sqrt{\alpha} t) + l_0$

Maximum Deviation of Stretch from Equilibrium Value:

$\alpha = (\frac{k}{m_1} + \frac{k}{m_2}) = (\frac{50}{0.1} + \frac{50}{0.3}) = \frac{2000}{3} \\ \text{max stretch} = \sqrt{\frac{k}{\alpha m_2}} (l_0 - x_{2_0}) = \sqrt{\frac{50}{\frac{2000}{3} \times 0.3}} \, (0.08 - 0.06) = 0.01 = \boxed{1 \text{cm}}$

During the initial period before the spring reaches its equilibrium length for the first time, the wall applies a net impulse to the system, which causes it to translate. Therefore, the answer to the first part of the question is also "1".