In the figure a beam supported by 2 springs with a different stiffness. Gravitation and friction can be neglected. The position and attitude of the system is defined by coordinates $x_1$ and $x_2$ and $sin(\theta)$=$\theta$.
If I'm correct the vertical motion of the center of mass of this particular system is defined as:
\begin{equation*}
m\ddot x_1+m\ddot x_2=-k_{1}x_{1}-k_{2}x_{2}
\end{equation*}
\begin{equation*}
m\ddot x_1+m\ddot x_2+k_{1}x_{1}+k_{2}x_{2}=0
\end{equation*}
\begin{equation}
\ddot x_1+\ddot x_2+\frac{k_{1}}{m}x_{1}+\frac{k_{2}}{m}x_{2}=0 \label{vert}
\end{equation}
And what is then the equation for the rotation? Would that then be:
\begin{equation}
\ddot\theta_1\frac{b}{2}+\ddot\theta_2\frac{b}{2}+\frac{k_{1}}{m}\theta_1\frac{b}{2}+\frac{k_{2}}{m}\theta_2\frac{b}{2}=0 \label{rot}
\end{equation}
Thanks in advance!
Best Answer
Consider the center of mass at $c= \frac{b}{2}$ and its displacement $$x_c = \frac{x_1+x_2}{2}$$ The slope of the body is $$\theta = \frac{x_2-x_1}{b}$$
These two are the degrees of freedom of the body. The equations are transformed to use these variables with $$\begin{align} x_1 & = x_c - \frac{b}{2} \theta \\ x_2 &= x_c + \frac{b}{2} \theta \end{align}$$
The force on each spring is $F_1 =-k_1 x_1$ and $F_2 =-k_2 x_2$. The equations of motion are derived from the sum of forces on body and sum of moments on the center of mass
$$ \left. \begin{aligned} F_1 + F_2 &= m \ddot{x}_c \\ \frac{b}{2} F_2 - \frac{b}{2} F_1 & = I \ddot{\theta} \end{aligned} \right\} \begin{aligned} -(k_1+ k_2) x_c - \frac{b}{2}(k_2-k_1) \theta &= m \ddot{x}_c \\ -\frac{b}{2} \left( (k_2-k_1) x_c + \frac{b}{2} (k_1+k_2) \theta \right) & = I \ddot{\theta} \end{aligned} $$
Now collect the terms to get the differential equation
$$ \begin{pmatrix} \ddot{x}_c \\ \ddot{\theta} \end{pmatrix} = - \begin{vmatrix} \frac{k_1+k_2}{m} & \frac{ \frac{b}{2} (k_2-k_1)}{m} \\ \frac{ \frac{b}{2} (k_2-k_1)}{I} & \frac{ \left( \frac{b}{2} \right)^2 (k_1+k_2)}{I} \end{vmatrix} \begin{pmatrix} x_c \\ \theta \end{pmatrix} $$