nondimensionalize the two mass system Consider the following two mass system executing rectilinear motion: The first mass, M1,
is connected to the left wall by a nonlinear spring with force law: $F_1(x) = −kx − αx^3$
.
A linear spring, with force law, $F_2(x) = −kx$ connects the large second mass, M2 to the
first mass, initially to the right of the first mass. Assume the initial conditions are that
the first mass is initially displaced an amount H with no initial velocity and the second
mass is not displaced and not moving initially.
(a) Write down the coupled second order, nonlinear differential equations governing this
system.
(b) Non-dimensionalize the problem so that a small parameter $\epsilon =\frac{M_1}{M_2}$
emerges and identify the second non-dimensional parameter, δ on the nonlinearity.
My thoughts: (a)\begin{equation}
\left\{\begin{split}
F_1(H) -F_2(0-H) = M_1 \frac{d^2 x_1}{dt^2} \\
F_2(0-H) = M_2 \frac{d^2 x_2}{dt^2} \\ 
\end{split}\right.
 \end{equation}
so I get
\begin{equation}
\left\{\begin{split}
-2KH - α H^3 = M_1 \frac{d^2 x_1}{dt^2} \\
-KH = M_2 \frac{d^2 x_2}{dt^2} \\ 
\end{split}\right.
 \end{equation}
with $x_1(0) = H$, $x_2(0) = 0$, $x_1'(0) = x_2'(0) = 0$.
(b) let $x_1 = b \hat{x_1}$, $x_2 = c \hat{x_2}$, $t = d \hat{t}$
then \begin{equation}
\left\{\begin{split}
-2KH - α H^3 = M_1 \frac{b}{d^2} \frac{d^2 \hat{x_1}}{dt^2} \\
-KH = M_2 \frac{c}{d^2} \frac{d^2 \hat{x_2}}{dt^2} \\ 
\end{split}\right.
 \end{equation}
which is equivalent to
\begin{equation}
\left\{\begin{split}
-2 - α \frac{H^2}{K} = M_1 \frac{b}{d^2KH} \frac{d^2 \hat{x_1}}{dt^2} \\
-1 = M_2 \frac{c}{d^2KH} \frac{d^2 \hat{x_2}}{dt^2} \\ 
\end{split}\right.
 \end{equation}
with $\hat{x_1}(0) = \frac{H}{b}$, $\hat{x_2}(0) = 0$, $\hat{x_1}'(0) = \hat{x_2}'(0) = 0$
so I set $\frac{H}{b} = 1$, $M_1 \frac{b}{d^2KH} = 1$ and $M_2 \frac{c}{d^2KH} = 1$ by set $b = H$, $c = \frac{HM_1}{M_2}$ and $d = \sqrt{\frac{M_1}{K}}$,
then I have the system :
\begin{equation}
\left\{\begin{split}
-2 - α \frac{H^2}{K} = \frac{d^2 \hat{x_1}}{dt^2} \\
-1 = \frac{d^2 \hat{x_2}}{dt^2} \\ 
\end{split}\right.
 \end{equation}
with $\hat{x_1}(0) = 1$, $\hat{x_2}(0) = 0$, $\hat{x_1}'(0) = \hat{x_2}'(0) = 0$
but I do not understand how to involve $\epsilon = \frac{M_1}{M_2}$ in the system and what is the second non-dimensional parameter δ on the nonlinearity. Thanks for any help!
 A: (a) The first mass is located at a distance $x_1$ from its equilibrium position, and similarly, the second mass is located at a distance $x_2$ from its equilibrium  position. This way, the elongation of the first spring w.r.t. equilibrium is $x_1$, whereas the elongation of the second spring is $x_2-x_1$. The force applied to the first mass is
$$
F_{\to 1} = F_1 - F_2 = -kx_1 - \alpha x_1^3 + k(x_2 - x_1) ,
$$
where the second mass is located at a distance $x_2$ from its equilibrium  position. Similarly,
$$
F_{\to 2} = F_2 = -k(x_2 - x_1) .
$$
Thus, Newton's laws yield
\begin{aligned}
M_1 \ddot x_1 &= -2kx_1 - \alpha x_1^3 + kx_2 ,\\
M_2 \ddot x_2 &= -k(x_2-x_1) ,
\end{aligned}
see this post for the linear case $\alpha=0$. At equilibrium, we find the values $x_2 = x_1 =0$ if $\alpha \geq 0$, which is implicitly assumed. Initial conditions are $$x_1(0) = H, \quad\dot x_1(0) = 0, \quad x_2(0)=0, \quad\dot x_2(0) = 0.$$
(b) We introduce dimensionless variables and coordinate
$$
\bar x_i(\bar t) = x_i(t) /H ,\qquad \bar t = t \sqrt{k/M_2}
$$
so that our system becomes
\begin{aligned}
\epsilon \bar x_1'' &= -2 \bar x_1 - \delta \bar x_1^3 + \bar x_2 ,\\
\bar x_2'' &= -(\bar x_2-\bar x_1) ,
\end{aligned}
with $\epsilon = M_1/M_2$ and $\delta = \alpha H^2/k$, and
$$\bar x_1(0) = 1, \quad \bar x_1'(0) = 0, \quad \bar x_2(0)=0, \quad \bar x_2'(0) = 0.$$
