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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.

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4 Answers 4

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.

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Well yes, I know that much. But I was hoping that you would help me with the reactions. Since the first ball isnt attached to a wall, the force for the first ball would be (0)x1 - k(x1 - x2 - x3). But Im not sure if that is right. The force of tue second and third ball would become like the two mass spring system attached to a wall, at least I think it will, so I get that, but I need help with the first. –  Additional Pylons Nov 4 '12 at 1:03
    
The coordinates are the positions of each ball. The forces are the changes in spring length, so the force on the first ball is $kx_1$ from the spring to the wall and $-k(x_2-x_1)$ for the spring to the second ball. –  Ross Millikan Sep 16 '13 at 0:29

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 $\mathbf{\Lambda}$.

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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.

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Infinite mass $M$ at the left and at the right

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.

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