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After solving a physics problem, this was the equation I came to:

$$ \left\{ \begin{array}{c} m_2\ddot x_2+k(x_2-x_1-l)=0 \\ m_1\ddot x_1 + m_2 \ddot x_2=0 \end{array} \right. $$

Where $m_1$ and $m_2$, the masses; $k$, the spring coefficient; and $l$, spring length are constants.

$x_1$ and $x_2$ are our variables.

If the two masses were identical, it could be easily transformed into the SHM equation and be solved using the substitution $u=x_2-x_1-l$.

But now, when I try to substitute, the equation will still contain other-than-u variables. That makes it impossible for me to solve.

Can anyone help me?

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  • $\begingroup$ Look up reduced mass. You'll need the substitutions $\frac{1}{\mu}=\frac{1}{m_1}+\frac{1}{m_2}$ and $x=x_1-x_2$. Note the second substitution can be differentiated. $\endgroup$ Commented Aug 17, 2016 at 8:18

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$$ \left\{ \begin{array}{c} m_2 x''_2+k(x_2-x_1-l)=0 \\ m_1 x''_1 + m_2 x''_2=0 \end{array} \right. $$

$x_1=\frac{m_2}{k}x''_2+x_2-l$

$x''_1=\frac{m_2}{k}x''''_2+x''_2 = -\frac{m_2}{m_1}x''_2$

$$m_1m_2x''''_2+k(m_1+m_2)x''_2 =0$$ This is a fourth order linear ODE. It can easily be reduced to second order and then be solved, leading to $x_2(t)$ .

Putting this result into $\quad x_1=\frac{m_2}{k}x''_2+x_2-l\quad $ gives $x_1(t)$.

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