(a) Write the differential equations satisfied by the principal coordinates for the system of Example 8.2. (b) Find the relation between the principal coordinates and the original generalized coordinates and vice versa. (c) Motion of the system is initiated by moving the third mass a distance δ

Question

(a) Write the differential equations satisfied by the principal coordinates for the system of Example 8.2.
(b) Find the relation between the principal coordinates and the original generalized coordinates and vice versa.
(c) Motion of the system is initiated by moving the third mass a distance [latex]\delta[/latex] from equilibrium while holding the other masses in their equilibrium position and then releasing the system from rest. Solve for the response of the principal coordinates.

Accepted Answer

(a) Recalling from Example 8.2, the natural frequencies of the system are

[latex]\omega _{1}=0.893\sqrt{\frac{k}{m} } [/latex] [latex]\omega _{2}=2.110\sqrt{\frac{k}{m} } [/latex] [latex]\omega _{3}=2.597\sqrt{\frac{k}{m} } [/latex] (a)

The differential equations governing the principal coordinates are

[latex] \ddot{p}_{1}+\left(0.893\sqrt{\frac{k}{m} } ight)^{2}p_{1}=0 [/latex] (b)

[latex]\ddot{p}_{2}+\left(2.110\sqrt{\frac{k}{m} } ight)^{2}p_{2}=0 [/latex] (c)

[latex]\ddot{p}_{3}+\left(2.597\sqrt{\frac{k}{m} } ight)p_{3}=0 [/latex] (c)

(b) The normalized eigenvectors are calculated in Example 8.10 as

[latex]\pmb{X}_{1}=\frac{1}{\sqrt{m} }\begin{bmatrix}0.659 \ 0.726 \ 0.279 \end{bmatrix} [/latex] [latex]\pmb{X}_{2}=\frac{1}{\sqrt{m} }\begin{bmatrix}-0.712 \ 0.518 \ 0.670 \end{bmatrix} [/latex] [latex]\pmb{X}_{1}=\frac{1}{\sqrt{m} }\begin{bmatrix}-0.242 \ 0.453 \ -1.213 \end{bmatrix} [/latex] (d)

he modal matrix is the matrix whose columns are the normalized eigenvectors

[latex]\pmb{P}=\frac{1}{\sqrt{m} }\begin{bmatrix} 0.659 & -0.712 & -0.242 \ 0.726 & 0.518 & 0.453 \ 0.279 & 0.670 & -1.213 \end{bmatrix} [/latex] (e)

The relation between the two sets of coordinates is given by Equation (8.74)

[latex] \pmb{X}=\pmb{Pp} \Rightarrow \begin{bmatrix} x_{1}(t)\x_{2}(t) \x_{3}(t)\end{bmatrix}=\frac{1}{\sqrt{m} } \begin{bmatrix} 0.659 & -0.712 & -0.242 \ 0.726 & 0.518 & 0.453 \ 0.279 & 0.670 & -1.213\end{bmatrix} \begin{bmatrix} p_{1}(t) \ p_{2}(t)\p_{3}(t)\end{bmatrix} [/latex] (f)

[latex] \pmb{M \ddot{X}}+(\alpha K+\beta \pmb{M})\pmb{X}+\pmb{K X} = \pmb{0}[/latex] (8.74)

The relationship is inverted yielding

[latex] \pmb{p}=\pmb{P}^{-1}\pmb{_{X}} \Rightarrow\begin{bmatrix} p_{1}(t)\ p_{2}(t)\p_{3}(t)\end{bmatrix} =\sqrt{m}\begin{bmatrix}0.659 & 0.726 & 0.140 \ -0.712 & 0.518 & 0.335 \ -0.242 & 0.453& -0.607\end{bmatrix} \begin{bmatrix} x_{1}(t)\x_{2}(t)\x_{3}(t) \end{bmatrix} [/latex] (g)

(c) The initial conditions for [latex]\pmb{x}[/latex] are

[latex]\begin{bmatrix} x_{1}(0)\ x_{2}(0)\x_{3}(0) \end{bmatrix} =\begin{bmatrix} 0 \ 0 \\delta \end{bmatrix} [/latex] [latex]\begin{bmatrix} \dot{x}_{1}(0)\ \dot{x}_{2}(0)\\dot{x}_{3}(0) \end{bmatrix} = \begin{bmatrix} 0 \ 0 \0 \end{bmatrix} [/latex] (h)

The initial conditions for the principal coordinates are obtained from Equation (g) as

[latex] \begin{bmatrix} p_{1}(0)\ p_{2}(0)\p_{3}(0)\end{bmatrix}=\sqrt{m}\begin{bmatrix}0.659 & 0.726 & 0.140 \ -0.712 & 0.518 & 0.335 \ -0.242 & 0.453 & -0.607 \end{bmatrix} \begin{bmatrix} 0 \ 0 \\delta \end{bmatrix}= \sqrt{m}\delta\begin{bmatrix}0.140 \ 0.335 \ -0.607 \end{bmatrix} [/latex] (i)

and

[latex] \begin{bmatrix} \dot{p}_{1}(0) \ \dot{p}_{2}(0)\ \dot{p}_{3}(0) \end{bmatrix}=\sqrt{m}\begin{bmatrix} 0.659 &0.726 & 0.140 \ -0.712 & 0.518 & 0.335 \-0.242 & 0.453 & -0.607 \end{bmatrix}\begin{bmatrix} 0\0\0 \end{bmatrix}= \begin{bmatrix} 0\0\0 \end{bmatrix} [/latex] (j)

The general solution for the principal coordinates is

[latex]p_{1}(t)=A_{1}\sin\left(0.893\sqrt{\frac{k}{m} }t ight) +B_{1}\cos\left(0.893\sqrt{\frac{k}{m} } ight) [/latex] (k)

[latex]p_{2}(t)=A_{2}\sin\left(2.1\sqrt{\frac{k}{m} } t ight)+B_{2}\cos \left(2.110\sqrt{\frac{k}{m} }t ight) [/latex] (l)

[latex] p_{3}(t)=A_{3}\sin\left(2.597\sqrt{\frac{k}{m} }t ight)+B_{3}\cos\left(2.597\sqrt{\frac{k}{m}t } ight) [/latex] (m)

Application of initial conditions, Equations (i) and (j) lead to [latex]B_{1}=0.140\delta \sqrt{m}, B_{2}=0.335\delta \sqrt{m},B_{3}=-0.607\delta \sqrt{m} ,A1=0, A_{2}=0 [/latex], and [latex]A_{3}=0 [/latex]. The original generalized coordinates are obtained using Equation (f) as

[latex]x_{1}(t)=0.0922\delta \cos\left(0.893\sqrt{\frac{k}{m} }t ight) -0.238\delta \cos\left(2.110\sqrt{\frac{k}{m} }t ight) [/latex]
[latex]+0.147\delta \cos\left(2.597\sqrt{\frac{k}{m} }t ight) [/latex] (n)

[latex]x_{2}(t)=0.102\delta \cos\left(0.893\sqrt{\frac{k}{m} }t ight)+0.174\delta \cos\left(2.110\sqrt{\frac{k}{m} }t ight) [/latex]
[latex] -0.275\delta \cos\left(2.597\sqrt{\frac{k}{m} }t ight) [/latex] (o)

[latex]x_{3}(t)=0.0389\delta \cos\left(0.893\sqrt{\frac{k}{m} }t ight)+0.224\delta\left(2.110\sqrt{\frac{k}{m} }t ight) [/latex]
[latex] +0.736\delta \cos\left(2.597\sqrt{\frac{k}{m} }t ight) [/latex] (p)

Equations (n) through (p) are the same as Equation (e) of Example 8.4.

Question 8.12: (a) Write the differential equations satisfied by the princi...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions