Modes of Vibration of 3-DOF Spring Mass System

[Reefy]

Homework Statement

Homework Equations

a₁₁ = a₂₁ = a₃₁ = a₁₂ = a₁₃

a₂₂ = a₃₂ = a₂₃

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

$$x_{1} = a_{11}mω^2x_{1}+a_{12}mω^2x_{1}+a_{13}mω^2x_{1}$$
$$x_{2} = a_{21}mω^2x_{2}+a_{22}mω^2x_{2}+a_{23}mω^2x_{2}$$
$$x_{3} = a_{31}mω^2x_{3}+a_{32}mω^2x_{3}+a_{33}mω^2x_{3}$$

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.

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

 

[Dr.D]

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

 

[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
X_1,2,3 = Asinωt, Bsinωt, and Csinωt are

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

 

[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.

 

[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

 

[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

 

[Dr.D]

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