Linearised 2-DOF Vibration Analysis

Let the coordinate system's origin be placed at the point of attachment of the spring with the wall. X-axis is horizontal to the right and Y axis is vertically up. The coordinates of the point masses are:

$x_1 = x$ $y_1 = 0$ $x_2 = x + L\sin{\theta}$ $y_2 = -L\cos{\theta}$

$\theta$ is the angle the rod makes with the vertical. Now, the kinetic and potential energies of the system are:

$T = \frac{m_1}{2} \left(\dot{x}_1^2 + \dot{y}_1^2\right) + \frac{m_2}{2} \left(\dot{x}_2^2 + \dot{y}_2^2\right)$ $T = \frac{(m_1+m_2)\dot{x}^2}{2} + m_2L\dot{x}\dot{\theta} \cos{\theta} + \frac{m_2 L^2 \dot{\theta}^2}{2}$

$V = \frac{K}{2}\left(x - L_o\right)^2 - m_2gL\cos{\theta}$

Since there are no dissipative forces acting on the system, the total energy of the system is conserved. The total energy is constant and therefore, its time derivative is zero.

$E = T+V$ $\frac{dE}{dt} = \left(f_1(x,\dot{x},\ddot{x},\theta,\dot{\theta},\ddot{\theta})\right)\dot{x} + \left(f_2(x,\dot{x},\ddot{x},\theta,\dot{\theta},\ddot{\theta})\right)\dot{\theta} =0$

Where:

$f_1(x,\dot{x},\ddot{x},\theta,\dot{\theta},\ddot{\theta}) = (m_1+m_2)\ddot{x} +m_2L\cos{\theta} \ddot{\theta} -m_2L\sin{\theta}\dot{\theta}^2 + K(x-L_o)$ $f_2(x,\dot{x},\ddot{x},\theta,\dot{\theta},\ddot{\theta}) =m_2L\cos{\theta} \ddot{x} + m_2L^2 \ddot{\theta} + m_2gL\sin{\theta}$

Now, the only way for the time derivative of energy to be zero for a system undergoing motion is if: $f_1(x,\dot{x},\ddot{x},\theta,\dot{\theta},\ddot{\theta}) = (m_1+m_2)\ddot{x} +m_2L\cos{\theta} \ddot{\theta} -m_2L\sin{\theta}\dot{\theta}^2 + K(x-L_o)=0$ $f_2(x,\dot{x},\ddot{x},\theta,\dot{\theta},\ddot{\theta}) =m_2L\cos{\theta} \ddot{x} + m_2L^2 \ddot{\theta} + m_2gL\sin{\theta}=0$

If $\dot{x}=0$ and $\dot{\theta}=0$ , that is the trivial 'no-motion' solution which is of no interest to us. Thus, the equations of motion are obtained. The equations can also conveniently be found using Lagrangian or Hamiltonian mechanics.

Having found the equations of motion, and with a little bit of intuition, one can conclude that when $x=L_o$ , $\dot{x}=\ddot{x}=0$ and $\theta=\dot{\theta}=\ddot{\theta}=0$ , the system is in stable equilibrium. The next step is to linearise this system about this equilibrium point. Let the system be displaced slightly such that $x = L_o + x_L$ and $\theta = 0 + \theta_L$ where $x_L$ and $\theta_L$ are small in magnitude. Then, the following Taylor expansion can be performed:

$f_1(x,\dot{x},\ddot{x},\theta,\dot{\theta},\ddot{\theta}) \approx f_1(L_o,0,0,0,0,0) + \left(\frac{\partial f_1}{\partial x} \biggr\rvert_{\mathrm{eq}}x_L\right) + \left(\frac{\partial f_1}{\partial \theta} \biggr\rvert_{\mathrm{eq}}\theta_L\right) +\left( \frac{\partial f_1}{\partial \dot{x}} \biggr\rvert_{\mathrm{eq}}\dot{x}_L\right)+\left( \frac{\partial f_1}{\partial \dot{\theta}} \biggr\rvert_{\mathrm{eq}}\dot{\theta}_L\right)+ \left(\frac{\partial f_1}{\partial \ddot{x}} \biggr\rvert_{\mathrm{eq}}\ddot{x}_L\right)+ \left(\frac{\partial f_1}{\partial \ddot{\theta}} \biggr\rvert_{\mathrm{eq}}\ddot{\theta}_L\right)$

Here, all derivatives at evaluated at the equilibrium point which corresponds to $x=L_o$ , $\dot{x}=\ddot{x}=0$ and $\theta=\dot{\theta}=\ddot{\theta}=0$ . After performing these evaluations and doing the same for $f_2$ gives the linearised equations of motion which are:

$f_1(x,\dot{x},\ddot{x},\theta,\dot{\theta},\ddot{\theta}) \approx (m_1+m_2)\ddot{x}_L + m_2L\ddot{\theta}_L + Kx_L=0$ $f_2(x,\dot{x},\ddot{x},\theta,\dot{\theta},\ddot{\theta}) \approx m_2L\ddot{x}_L + m_2L^2\ddot{\theta}_L + m_2gL\theta_L=0$

The equations of motion can be rearranged in a matrix form as such:

$\left[\begin{matrix} m_1+m_2&m_2L\\m_2L&m_2L^2\end{matrix}\right] \left[\begin{matrix} \ddot{x}_L\\ \ddot{\theta}_L\end{matrix}\right]+ \left[\begin{matrix} K&0\\0&m_2gL\end{matrix}\right] \left[\begin{matrix} x_L\\ \theta_L\end{matrix}\right]=\left[\begin{matrix}0\\0\end{matrix}\right]$

$\bar{M} \ddot{q} + \bar{K} q = 0$

Thus, the linearised equations of motion are written in the above form. Here, $\bar{M}$ and $\bar{K}$ are known as the mass and stiffness matrix respectively. Now, let us assume the solution to be of the following form:

$q =A \mathrm{v} \sin(\omega t + \phi)$

Here $v$ is an unknown column matrix of length 2 and $A$ is an unknown scalar. Plugging these values into the linearised system of equations of motion:

$-\omega^2 \bar{M} \mathrm{v} A\sin(\omega t + \phi) + \bar{K} \mathrm{v} A\sin(\omega t + \phi) =0$ $-\omega^2 \bar{M} \mathrm{v} + \bar{K} \mathrm{v}=0$

$(\bar{M}^{-1}\bar{K}) \mathrm{v}= \omega^2 \mathrm{v}$ This is an eigenvalue problem, the eigenvalues of which are $\omega^2$ . Finding the characteristic equation and solving for the sum of eigenvalues gives us the required answer of $\boxed{22.5}$ after plugging in all parameters. This method can be used to solve linear vibration problems as well. I have posted problems on such systems earlier.