Question

6.84 Using modal analysis, determine the free-vibration response of a two-degree-of-freedom system with equations of motion $2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \ddot{\vec{x}}(t) + 8 \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \vec{x}(t) = \vec{0}$ with initial conditions $\vec{x}(0) = \begin{Bmatrix} 1 \\ 0 \end{Bmatrix}, \dot{\vec{x}}(0) = \begin{Bmatrix} 0 \\ 1 \end{Bmatrix}$

Name: 684 using modal analysis determine the free vibration response of a two degree of freedom system with equations of motion 2beginbmatrix 1 0 0 1 endbmatrix ddotvecxt 8 beginbmatrix 2 1 1 2 en 59097
Uploaded: 2023-02-01T18:10:53-08:00
Duration: 2 min 44 s
Channel: Sam Stansfield
Description: 684 using modal analysis determine the free vibration response of a two degree of freedom system with equations of motion 2beginbmatrix 1 0 0 1 endbmatrix ddotvecxt 8 beginbmatrix 2 1 1 2 en 59097

          6.84 Using modal analysis, determine the free-vibration response of a two-degree-of-freedom
system with equations of motion

$2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \ddot{\vec{x}}(t) + 8 \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \vec{x}(t) = \vec{0}$

with initial conditions

$\vec{x}(0) = \begin{Bmatrix} 1 \\ 0 \end{Bmatrix}, \dot{\vec{x}}(0) = \begin{Bmatrix} 0 \\ 1 \end{Bmatrix}$

684 using modal analysis determine the free vibration response of a two degree of freedom system with equations of motion 2beginbmatrix 1 0 0 1 endbmatrix ddotvecxt 8 beginbmatrix 2 1 1 2 en 59097

Added by Brandy M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sam Stansfield

02/01/2023

Step 1

Step 1: First, we can simplify the equation of motion as: $M\ddot{\vec{x}}(t) + K\vec{x}(t) = \vec{0}$ where $M = 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $K = 8\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$. Show more…

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6.84 Using modal analysis, determine the free-vibration response of a two-degree-of-freedom system with equations of motion $2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \ddot{\vec{x}}(t) + 8 \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \vec{x}(t) = \vec{0}$ with initial conditions $\vec{x}(0) = \begin{Bmatrix} 1 \\ 0 \end{Bmatrix}, \dot{\vec{x}}(0) = \begin{Bmatrix} 0 \\ 1 \end{Bmatrix}$