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## Homework Statement

A system composed of two blocks of mass ##m_1## and ##m_2## connected by a massless spring with spring constant ##k##. The blocks slide on a frictionless plane. The unstretched length of the spring is ##l##. Initially, ##m_2## is held so that the spring is compressed at ##\frac{l}{2}## and ##m_1## is forced against a stop. ##m_2## is released at ##t=0##.

Find the motion of the center of mass.

## Homework Equations

Spring restoration force, center of mass.

## The Attempt at a Solution

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__First part: mass 1 does not move__

Their is a time ##0\le t\le t_m## where the mass ##m_1## will be stuck to the wall, and the external force from the wall on the system will be the spring restoration force, so ##f_{ext} = k(l-x_2)##.

Since ## x_1(t) = 0 ##, the center of mass is ## x_{com}(t) = \frac{m_2}{m_1+m_2} x_2(t) ##.

We must therefore solve the differential equation

## (m_1 + m_2) \ddot x_{com} = f_{ext} \iff m_2 \ddot x_2 = k(l-x_2) ##,

which is not homogenous, so we put ## u = x_2 - l ## and solve ## \ddot u + \frac{k}{m_2} u = \ddot u + \omega^2 u = 0##.

With initial conditions ##u(0) = -\frac{l}{2}## and ##\dot u(0) = 0 ##, we find that:

## x_{com}(t) = \frac{m_2l}{m_1+m_2}(1-\frac{\cos(\omega t)}{2})##

Now the maximum stretch before mass 1 moves is attained when ## \omega t_m = \frac{\pi}{2} ## so ## t_m = \frac{\pi}{2} \sqrt{\frac{m_2}{k}}##

__Second part: mass 1 moves but there are no more external forces.__

Therefore the center of mass must have constant speed for ## t\ge t_m##, so its motion must be :

##x_{com}(t) = x_{com}(t_m) + (t-t_m) v(t_m) = \frac{m_2l}{m_1+m_2}(1-\frac{\pi}{4}+\frac{t\omega}{2})##

Do you think it is the right solution?

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