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Math Help - mechanics:please help

  1. #1
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    mechanics:please help

    Two degree-of-freedom systems
    Find the natural frequencies and describe the modes for the linear vibration system below.
    Attached Thumbnails Attached Thumbnails mechanics:please help-2nddf.jpg  
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  2. #2
    Grand Panjandrum
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    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
    Last edited by CaptainBlack; August 25th 2006 at 10:08 PM.
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  3. #3
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    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:

    <br />
{m_1 \ddot{x}=-k_1 x+k_2(x-y)} \atop<br />
{m_2 \ddot{y}=-k_3 y-k_2(x-y)}<br />

    Now substitute in the given values:

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

    RonL
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  4. #4
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    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:

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

    which may be rewritten as:

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

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

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

    where:

    <br />
\bold{A}=<br />
\left [ \begin{array}{cc}<br />
{-73.333}& {6.667}\\<br />
{6}& {-80}<br />
\end{array} \right]<br />

    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:

    <br />
{0.867895 \brack 0.496748}<br />
and <br />
{-0.536649 \brack 0.843805}<br />
.

    RonL

    I will complete the last part of this tommorow, its midnight here now and I
    am going to bed.
    Last edited by CaptainBlack; August 25th 2006 at 02:43 PM.
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  5. #5
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    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:

    <br />
\bold{X}={x \brack y}<br />

    in terms of the eigen vectors, so we put:

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

    where:

    <br />
\bold{U}={u_1 \brack u_2}<br />
,

    and:

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

    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}} <br />
=\bold{AX}=\bold{ACU}<br />
.

    Hence:

    <br />
\ddot{\bold{U}}=\{ \bold{C}^{-1}\bold{AC}\} \ \bold{U}<br />
,

    and:

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

    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
    Last edited by CaptainBlack; August 25th 2006 at 10:09 PM.
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  6. #6
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    Quote Originally Posted by CaptainBlack
    Hence:

    <br />
\ddot{\bold{U}}=\{ \bold{C}^{-1}\bold{AC}\} \ \bold{U}<br />
,

    and:

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

    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:

    <br />
\ddot{\bold{U}}=\bold{B} \ \bold{U}<br />
,

    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
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  7. #7
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    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..?
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  8. #8
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    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
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  9. #9
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    here is the characteristic equation i got..
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  10. #10
    Grand Panjandrum
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    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
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