# Homework Help: Motion of c.o.m. (kleppner 3.7)

1. Dec 11, 2014

### geoffrey159

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Spring restoration force, center of mass.

3. The attempt at a solution

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?

Last edited: Dec 11, 2014
2. Dec 11, 2014

### haruspex

You can get to this a bit faster. Up to here, we could treat m1 as fixed to the stop. Going from max compression to zero will be one quarter period.
Doesn't seem right. Please post the individual steps.

3. Dec 11, 2014

### geoffrey159

Ok so in details, I get :
$x_{com}(t_m) = \frac{m_2l}{m_1+m_2}$
$v(t_m) = \dot x_{com}(t_m) = \frac{m_2l}{m_1+m_2} \frac{\omega}{2}$

So in the end, for $t\ge t_m$ :
$x_{com}(t) = \frac{m_2l}{m_1+m_2} (1+(t-t_m)\frac{\omega}{2}) = \frac{m_2l}{m_1+m_2}(1-\frac{\pi}{4}+\frac{t\omega}{2})$

Last edited: Dec 11, 2014
4. Dec 11, 2014

### haruspex

Yes, that looks right, except that you probably should not leave an $\omega$ in the answer (which is why I made the mistake of thinking we had different answers).