Two degree-of-freedom systems

Find the natural frequencies and describe the modes for the linear vibration system below.

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- Aug 25th 2006, 10:48 AMbobby77mechanics:please help
Two degree-of-freedom systems

Find the natural frequencies and describe the modes for the linear vibration system below. - Aug 25th 2006, 12:13 PMCaptainBlackQuote:

Originally Posted by**bobby77**

1. Decide on your variables (use the deflection of the first trolley from its

equilibrium position as x, and the second trolley from its equilibrium position

as y, positive to the right).

2. Write out the equations for the accelerations of the trolleys in terms of

x and y.

3. Rewrite as a vector equation X''=AX', where A is now a 2x2 matrix, and

X=[x,y]'.

4. Now the normal modes correspond to the eigen vectors of A.

5. If we now rewrite the equation in 3 in terms of the normal modes the

new A will be diagonal (with the eigen values of the original A on the

diagonal), and the solution to the DE is:

Y=exp(sqrt(A)t)

which will give the natural frequencies.

RonL - Aug 25th 2006, 01:06 PMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

$\displaystyle F_1=-k_1 x+k_2(x-y)$,

and on trolly 2:

$\displaystyle F_2=-k_3 y-k_2(x-y)$,

where the notation is as in the attached figure.

So:

$\displaystyle

{m_1 \ddot{x}=-k_1 x+k_2(x-y)} \atop

{m_2 \ddot{y}=-k_3 y-k_2(x-y)}

$

Now substitute in the given values:

$\displaystyle

\begin{array}{ccc}

{\ddot{x}\ =}& {-73.333x}&{+6.667y}\\

{\ddot{y}\ =}& {6x}&{-80y}

\end{array}

$

RonL - Aug 25th 2006, 01:39 PMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

$\displaystyle

\begin{array}{ccc}{\ddot{x}\ =}& {-73.333x}&{+6.667y}\\{\ddot{y}\ =}& {6x}&{-80y}\end{array}

$

which may be rewritten as:

$\displaystyle

\left [ \begin{array}{c}

{\ddot{x}}\\

{\ddot{y}}

\end{array} \right]=$$\displaystyle

\left [ \begin{array}{cc}

{-73.333}& {6.667}\\

{6}& {-80}

\end{array} \right]$$\displaystyle

\left [ \begin{array}{c}

{x}\\

{y}

\end{array} \right]

$.

So writing $\displaystyle \bold{X}={x \brack y}$, this becomes:

$\displaystyle \ddot{\bold{X}}=\bold{AX}$,

where:

$\displaystyle

\bold{A}=

\left [ \begin{array}{cc}

{-73.333}& {6.667}\\

{6}& {-80}

\end{array} \right]

$

Now we may find the eigen values and vectors of $\displaystyle \bold{A}$ by hand, but I will resort to numerical packages at this point:

Code:

>A=[-73.333,6.667;6,-80]

-73.333 6.667

6 -80

>help eigen

function eigen (A)

## eigen(A) returns the eigenvectors and a basis of eigenvectors of A.

## {l,x,ll}=eigen(A), where l is a vector of eigenvalues,

## x is a basis of eigenvectors,

## and ll is a vector of distinct eigenvalues.

>

>{aa,bb,cc}=eigen(A)

-69.5171+0i -83.8159+0i

>aa

-69.5171+0i -83.8159+0i

>bb

0.867895+0i -0.536649+0i

0.496748+0i 0.843805+0i

>cc

-69.5171+0i -83.8159+0i

>

$\displaystyle

{0.867895 \brack 0.496748}

$ and $\displaystyle

{-0.536649 \brack 0.843805}

$.

RonL

I will complete the last part of this tommorow, its midnight here now and I

am going to bed. - Aug 25th 2006, 09:58 PMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

$\displaystyle

\bold{X}={x \brack y}

$

in terms of the eigen vectors, so we put:

$\displaystyle

\bold{X}={x \brack y}$$\displaystyle

=

{0.867895 \brack 0.496748}u_1

$$\displaystyle +

{-0.536649 \brack 0.843805}u2

$$\displaystyle =

\left [ \begin{array}{cc}

{0.867895}& {-0.536649}\\

{0.496748}& {0.843805}

\end{array} \right]{u_1 \brack u_2}

$$\displaystyle =\bold{CU}

$

where:

$\displaystyle

\bold{U}={u_1 \brack u_2}

$,

and:

$\displaystyle

\bold{C}=\left [ \begin{array}{cc}

{0.867895}& {-0.536649}\\

{0.496748}& {0.843805}

\end{array} \right]

$

Also:

$\displaystyle \bold{U}=\bold{C}^{-1}\bold{X}$.

Now we rewrite the differential equation in terms of $\displaystyle \bold{U}$:

$\displaystyle \ddot{\bold{X}}=\bold{C \ddot{U}}$$\displaystyle

=\bold{AX}=\bold{ACU}

$.

Hence:

$\displaystyle

\ddot{\bold{U}}=\{ \bold{C}^{-1}\bold{AC}\} \ \bold{U}

$,

and:

$\displaystyle \bold{C}^{-1}\bold{AC}=$$\displaystyle

\left [ \begin{array}{cc}{-69.5171}& {0}\\{0}& {-83.8159

}\end{array} \right]

$.

So the differential equations for $\displaystyle u_1$, and $\displaystyle u_2$

are uncoupled, and of the form which give simple harmonic motion, so the

usual methods can now be applied to obtain the natural frequencies.

RonL - Aug 25th 2006, 10:49 PMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

$\displaystyle

\ddot{\bold{U}}=\bold{B} \ \bold{U}

$,

then:

$\displaystyle \bold{U}=\exp(\sqrt{\bold{B}}t)$,

where what this means when $\displaystyle \bold{B}$ is diagonal is obvious (and I don't propose

to go into what conditions we need on $\displaystyle \bold{B}$ for this to work in general

or what it means for non-diagonal matrices).

RonL - Aug 27th 2006, 12:55 PMbobby77
according to the file attached..

I get characterstic equation as:

7500s^4+1150s^2+43=0

where s=natural frequency

solving this equation gives s=+/- 0.3i and s= +/-0.25i

is it correct natural frequency..? - Aug 28th 2006, 05:06 AMCaptainBlackQuote:

Originally Posted by**bobby77**

equation, and how you found it.

RonL - Aug 28th 2006, 08:59 AMbobby77
here is the characteristic equation i got..

- Aug 28th 2006, 09:03 AMCaptainBlackQuote:

Originally Posted by**bobby77**

kN/m rather than converted them to N/m.

RonL