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## Homework Statement:

- Derive equations of motion in terms of the displacements [itex] x_1 [/itex] and [itex] x_2 [/itex]

## Relevant Equations:

- F = ma

Hi,

So the question is to:

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?

I have decided to define an angle [itex] \theta [/itex] in the CCW direction from the point where the 2m mass is so that [itex] \theta = \frac{x_1 - x_2}{3L} [/itex]

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:

[tex] -k x_2 - k \left( x_2 + 2L \theta \right) = m \left( \ddot x_2 + L \ddot \theta \right) [/tex]

rearranging and substituting in for [itex] \ddot \theta [/itex] leads to:

[tex] m \ddot x_1 + 2m \ddot x_2 + 2k x_1 + 4k x_2 = 0 [/tex]

Because the bar is assumed to be rigid, I thought that the angle [itex] \theta [/itex] will be the same at the COM as it is at the right side

Taking moments, I get:

[tex] k x_2 L - 2mgL + mg(2L) - k \theta L^2 = I \ddot \theta [/tex]

noting that: [itex] I = m(2L)^2 + 2m(L)^2 = 6mL^2 [/itex] and rearranging we get:

[tex] 6m \ddot x_1- 6m \ddot x_2 + 4k x_1 - 4k x_2 = 0 [/tex]

Combining these into matrix form, we get:

[tex] \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 [/tex]

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.

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)(NB.

**
**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 [itex] \theta [/itex] in the CCW direction from the point where the 2m mass is so that [itex] \theta = \frac{x_1 - x_2}{3L} [/itex]

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:

[tex] -k x_2 - k \left( x_2 + 2L \theta \right) = m \left( \ddot x_2 + L \ddot \theta \right) [/tex]

rearranging and substituting in for [itex] \ddot \theta [/itex] leads to:

[tex] m \ddot x_1 + 2m \ddot x_2 + 2k x_1 + 4k x_2 = 0 [/tex]

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

Taking moments, I get:

[tex] k x_2 L - 2mgL + mg(2L) - k \theta L^2 = I \ddot \theta [/tex]

noting that: [itex] I = m(2L)^2 + 2m(L)^2 = 6mL^2 [/itex] and rearranging we get:

[tex] 6m \ddot x_1- 6m \ddot x_2 + 4k x_1 - 4k x_2 = 0 [/tex]

Combining these into matrix form, we get:

[tex] \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 [/tex]

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.