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: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
The force from the springs on trolly 1 is: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
The system of ODE is: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:
So we have eigen values $\displaystyle -69.5171$ and $\displaystyle -83.8159$ with corresponding eigen vectors: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.
Now we want to write: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
Of course the matrix solution to this also works, which is, if: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