Three ball-spring system So here is a crazy  problem for you all. Imagine there is a system of three balls in a line. The first and last balls have a larger mass M and the middle ball is a smaller mass m. Inbetwen the two larger balls and the middle ball are springs with spring constant of k. Now I would like to model this confusing heap of a system and solve the resulting eigenvalue problem . Trouble is, I don't k ow where to start. Normally, I would set up a differential relation, but since none of the balls are connected to walls, I don't know what is moving where? The problem doesn't even mention an external force. Also, what aould the eigenvalue solution to this problem even mean? I can't seem to grasp the meaning behind this problem. Please help.
 A: I would use a Lagrangian formulation.  Use the Euler-Lagrange equation
$$\frac{d}{dt} \frac{\partial L}{\partial \dot{q_k}} = \frac{\partial L}{\partial q_k}$$
where $L = T-V$ is the difference between the kinetic and potential energy.  The $q_k$, $k \in \{1,2,3\}$, are the displacements of the balls from equilibrium positions.  It should be evident that
$$2 L=M \dot{q_1}^2 + m \dot{q_2}^2 + M \dot{q_3}^2 - k (q_1-q_2)^2-k(q_2-q_3)^2$$
From the E-L equation, we get, in matrix form:
$$\mathbf{\ddot{q}} + \mathbf{\Lambda} \cdot \mathbf{q} = 0$$
where $\mathbf{q}=\left ( q_1,q_2,q_3 \right )^T$ and
$$\mathbf{\Lambda} = \frac{k}{M} \left ( \begin{array} \\ 1 & -1 & 0 \\ -M/m & 2 M/m & -M/m \\ 0 & -1 & 1\end{array} \right )$$
The eigenfrequencies are given by the square root of the eigenvalues of $\mathbf{\Lambda}$.
A: There are no external forces involved. Write down Newton's second law $\ddot{x}=F/m$ for all three balls. These will be three coupled differential equations which you can then write in matrix form.
A: 
We can consider infinity $M$ mass to the left and to the right of the small mass $m$. $\Psi_{n}$ is the $n\!-$ball displacement from its equilibrium position. Whith the conditions ( there is not any force to the left of $n = -1$ and to the right of $n = 1$ ):
$$
\Psi_{-2}\left(t\right) = \Psi_{-1}\left(t\right)\,,
\qquad
\Psi_{1}\left(t\right) = \Psi_{2}\left(t\right)\,;
\qquad
\forall\ t
$$ 
the systems of three balls ( at the center ) do not feel the effect of the rest of the ball ( with $n \leq -2$ and $n \geq 2$ ). So, we solve the 'infinity problem' and as a byproduct we get the solutions to our problem
( $\Psi_{-1}\left(t\right)$, $\Psi_{0}\left(t\right)$ and
$\Psi_{1}\left(t\right)$ ). Another condition is ( $m$ satisfies Newton Second
Law ):
$$
m\,{\partial^{2}\Psi_{0}\left(t\right) \over \partial t^{2}}
=
k_{\rm H}\left\lbrack\Psi_{1}\left(t\right) - \Psi_{0}\left(t\right)\right\rbrack
-
k_{\rm H}\left\lbrack\Psi_{0}\left(t\right) - \Psi_{-1}\left(t\right)\right\rbrack\,,
\qquad
\forall\ t 
$$
$\color{#ff0000}{%
\mbox{In order to avoid any confusion with the wave number}\
\color{#000000}{k},
\mbox{we changed}\ \color{#000000}{k}\ \mbox{by}\ \color{#000000}{k_{H}}\ \mbox{( Hooke }\ \color{#000000}{k}\ { ).}}$ 
Normal mode solutions are:
\begin{align}
\Psi_{n}\left(t\right)
&\equiv
\left\lbrack
A_{-}\sin\left(nka\right) + B_{-}\cos\left(nka\right)
\right\rbrack
\sin\left(\omega_{k}t + \phi_{-}\right)\,,
\qquad
n \leq 0
\\[3mm]
\Psi_{n}\left(t\right)
&\equiv
\left\lbrack
A_{+}\sin\left(nka\right) + B_{+}\cos\left(nka\right)
\right\rbrack\sin\left(\omega_{k}t + \phi_{+}\right)\,,
\qquad
n \geq 0
\end{align}
 where $a$ is the distance between adjacent balls in equilibrium.
$\omega_{k} = 2\omega_{\rm H}\sin\left(ka/2\right)$.
$\omega_{\rm H} \equiv \left(k_{H}/M\right)^{1/2}$. Since
$\Psi_{\pm n}\left(t\right)$ must agree at $n = 0$, we conclude that
$B_{-} = B_{+} \equiv B$ and $\phi_{-} = \phi_{+} \equiv \phi$.
\begin{align}
&\mbox{Border conditions yield}
\\[3mm]
&-A_{-}\sin\left(2ka\right) + B\cos\left(2ka\right)
=
-A_{-}\sin\left(ka\right) + B\cos\left(ka\right)
\\[3mm]
&-A_{+}\sin\left(2ka\right) + B\cos\left(2ka\right)
=
-A_{+}\sin\left(ka\right) + B\cos\left(ka\right)
\end{align}
\begin{align}
&\\[5mm]
&\mbox{Newton Second Law yields ( see above )}
\\[3mm]
&
-\omega_{k}^{2}\,B
=
\omega_{\rm h}^{2}\left\lbrack%
\left(A_{+} - A_{-}\right)\sin\left(ka\right)
+
\left(A_{+} + A_{-}\right)\cos\left(ka\right)
-
2B
\right\rbrack
\end{align}
where $\omega_{\rm h} \equiv \left(k_{H}/m\right)^{1/2}$.
Those equations yield a equation for the possible values of $k$. Do it.
A: Just reference the balls to a stationary coordinate system.  Then you can denote the positions and velocities with respect to that system.  Second, draw a free body diagram for each ball and write it's associated f=ma equation.  Then model the sates:  $x_i$ as the $i$th position and $v_i$ as the $i$th velocity.  Then $\sum f_i = m_i \dot v_i$ and $v_i = \dot x_i$ yields two differential equations for each ball. This should yield a system of six differential equations which can then be put into matrix form.  You will then be able to determine the frequency and modes of oscillation by solving for the eigenvalues and eigenvectors respectively.
