• Aug 25th 2006, 11:48 AM
bobby77
Two degree-of-freedom systems
Find the natural frequencies and describe the modes for the linear vibration system below.
• Aug 25th 2006, 01:13 PM
CaptainBlack
Quote:

Originally Posted by bobby77
Two degree-of-freedom systems
Find the natural frequencies and describe the modes for the linear vibration system below.

The proceedure for dealing with this problem is:

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, 02:06 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
The proceedure for dealing with this problem is:

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.

The force from the springs on trolly 1 is:

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

and on trolly 2:

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

where the notation is as in the attached figure.

So:

$
{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:

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

RonL
• Aug 25th 2006, 02:39 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack

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.

The system of ODE is:

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

which may be rewritten as:

$
\left [ \begin{array}{c}
{\ddot{x}}\\
{\ddot{y}}
\end{array} \right]=$
$
\left [ \begin{array}{cc}
{-73.333}& {6.667}\\
{6}& {-80}
\end{array} \right]$
$
\left [ \begin{array}{c}
{x}\\
{y}
\end{array} \right]
$
.

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

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

where:

$
\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 $\bold{A}$ 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 >
So we have eigen values $-69.5171$ and $-83.8159$ with corresponding eigen vectors:

$
{0.867895 \brack 0.496748}
$
and $
{-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, 10:58 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
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.

Now we want to write:

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

in terms of the eigen vectors, so we put:

$
\bold{X}={x \brack y}$
$
=
{0.867895 \brack 0.496748}u_1
$
$+
{-0.536649 \brack 0.843805}u2
$
$=
\left [ \begin{array}{cc}
{0.867895}& {-0.536649}\\
{0.496748}& {0.843805}
\end{array} \right]{u_1 \brack u_2}
$
$=\bold{CU}
$

where:

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

and:

$
\bold{C}=\left [ \begin{array}{cc}
{0.867895}& {-0.536649}\\
{0.496748}& {0.843805}
\end{array} \right]
$

Also:

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

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

$\ddot{\bold{X}}=\bold{C \ddot{U}}$ $
=\bold{AX}=\bold{ACU}
$
.

Hence:

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

and:

$\bold{C}^{-1}\bold{AC}=$ $
\left [ \begin{array}{cc}{-69.5171}& {0}\\{0}& {-83.8159
}\end{array} \right]
$
.

So the differential equations for $u_1$, and $u_2$
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, 11:49 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Hence:

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

and:

$\bold{C}^{-1}\bold{AC}=$ $
\left [ \begin{array}{cc}{-69.5171}& {0}\\{0}& {-83.8159
}\end{array} \right]
$
.

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

Of course the matrix solution to this also works, which is, if:

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

then:

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

where what this means when $\bold{B}$ is diagonal is obvious (and I don't propose
to go into what conditions we need on $\bold{B}$ for this to work in general
or what it means for non-diagonal matrices).

RonL
• Aug 27th 2006, 01:55 PM
bobby77
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, 06:06 AM
CaptainBlack
Quote:

Originally Posted by bobby77
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..?

Tell us what you have for the matrix for which this is the charateristic
equation, and how you found it.

RonL
• Aug 28th 2006, 09:59 AM
bobby77
here is the characteristic equation i got..
• Aug 28th 2006, 10:03 AM
CaptainBlack
Quote:

Originally Posted by bobby77
here is the characteristic equation i got..

Looking at your matrix it looks like you have left the spring constants in
kN/m rather than converted them to N/m.

RonL