# Mechanical Vibrations - 2 DOF System - Rod with 2 springs

• Engineering

## Homework Statement:

Derive equations of motion in terms of the displacements $x_1$ and $x_2$

## Relevant Equations:

F = ma
Hi,

So the question is to: derive the equations of motion for the following in terms of x1 and x2? The bar is assumed to be light and rigid.

(NB.
I know I posted another vibrations problem earlier in which I tried to use an energy approach to get to the equations of motion. However, we haven't been taught these, so I believe we are expected to reach these solutions by using force methods)

I am just wondering whether there is a general method when approaching these 2-DOF vibration problems? (e.g. introduce another variable for equations, specific places to take moments about, etc). I was wondering whether my method seems correct?

1) Resolve vertically downwards
I have decided to define an angle $\theta$ in the CCW direction from the point where the 2m mass is so that $\theta = \frac{x_1 - x_2}{3L}$

Also, I have identified that the COM is L away from the 2m mass. I have also made small angle approximations for the angles.

Therefore, we get the following when we resolve vertically downwards and get:
$$-k x_2 - k \left( x_2 + 2L \theta \right) = m \left( \ddot x_2 + L \ddot \theta \right)$$
rearranging and substituting in for $\ddot \theta$ leads to:
$$m \ddot x_1 + 2m \ddot x_2 + 2k x_1 + 4k x_2 = 0$$

2) Take moments about the centre of mass CCW direction
Because the bar is assumed to be rigid, I thought that the angle $\theta$ will be the same at the COM as it is at the right side

Taking moments, I get:
$$k x_2 L - 2mgL + mg(2L) - k \theta L^2 = I \ddot \theta$$
noting that: $I = m(2L)^2 + 2m(L)^2 = 6mL^2$ and rearranging we get:
$$6m \ddot x_1- 6m \ddot x_2 + 4k x_1 - 4k x_2 = 0$$

Combining these into matrix form, we get:
$$\begin{bmatrix} m & 2m \\ 6m & -6m \end{bmatrix} \begin{bmatrix} \ddot x_1 \\ \ddot x_2 \end{bmatrix} + \begin{bmatrix} 2k & 4k \\ 4k & -4k \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0$$

Does this equation look correct?

A final question on this: if I was to go on from here to find the natural frequencies/ etc., I would presume that it is okay to leave the matrices in their current form? I only ask that as the few examples I have seen in lectures only have 'M' matrices which are diagonal, but I presume that is reflective of the respective situation rather than a mathematical reason?

Thanks in advance for the help.