# Modes of Vibration of 3-DOF Spring Mass System

1. Nov 19, 2017

### Reefy

1. The problem statement, all variables and given/known data

2. Relevant equations
a11 = a21 = a31 = a12 = a13

a22 = a32 = a23

$$\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} =ω^2m \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}$$

$$x_{1} = a_{11}mω^2x_{1}+a_{12}mω^2x_{1}+a_{13}mω^2x_{1}$$
$$x_{2} = a_{21}mω^2x_{2}+a_{22}mω^2x_{2}+a_{23}mω^2x_{2}$$
$$x_{3} = a_{31}mω^2x_{3}+a_{32}mω^2x_{3}+a_{33}mω^2x_{3}$$
3. The attempt at a solution

I have already completed part a. The natural frequencies are 0, 1 and square root of 3 rad/s.

I'm having difficulty finding the modes of vibration. I wanted to use the influence coefficient method where I select the left-most mass to undergo a unit force while keeping the other masses fixed. This would cause a deflection of the left-moss mass and give me my first influence coefficient a11. However, there is no spring to the left of this mass and I'm having trouble figuring out how to apply the influence coefficient method.

$$a_{22} = a_{32} = a_{23} = \frac {1}{k}$$
$$a_{33} = \frac {2}{k}$$
$$a_{11} = a_{21} = a_{31} = a_{12} = a_{13}= ???$$

Last edited: Nov 19, 2017
2. Nov 19, 2017

### Dr.D

If you are seriously constrained to this particular method, you had better talk with your teacher some more.

3. Nov 19, 2017

### Reefy

Is there another method that you know of? Can I set up a ratio of the amplitudes? My equations after assuming a solution of
X1,2,3 = Asinωt, Bsinωt, and Csinωt are

$$A(1-ω^2) - B = 0$$
$$B(2-ω^2) - A - C = 0$$
$$C(1-ω^2) - B = 0$$

4. Nov 19, 2017

### Dr.D

For a particular value of omega (0,1,root3), choose A=1 and solve for B and C. That will give you the mode vector for the particular omega value.

5. Nov 19, 2017

### Reefy

Ok, if I do that then I get A = C = 1, and then B will equal 1, 0, and -2 for the three different ω

6. Nov 19, 2017

### Reefy

Something seems off about that because if I say A = 1, then

$$A(1-ω^2) - B = 0$$ $$→$$ $$B=(1-ω^2)$$

And

$$C(1-ω^2) - B = 0$$ $$→$$ $$C = B/(1-ω^2) = (1-ω^2)/(1-ω^2) = 1$$

But I still have the equation for $$B(2-ω^2) - A - C = B(2-ω^2) - 2 = 0$$ $$→$$ $$B=2/(2-ω^2)$$ which will give me slightly different values for B

7. Nov 20, 2017

### Dr.D

For omega = 0, I get for A=1, then B = C = 1. No inconsistency there, I think.