Coupled oscillators,
jiggled from one end

A one-dimensional system of n masses connected by springs.  The far end of the last spring, the spring connected to the right  , is fixed.  The first spring, the spring connected to the left of , can be "jiggled" from the left with displacement . The system of equations to solve is :

where, for an example of 3 masses,
X = (x ) 1 x 2 x 3
A = ( k + k k ) 1 2 2 -------- -- m m 1 2 0 k k + k k 2 2 3 3 -- -------- -- m m m 2 2 3 k k + k 2 3 4 -- -------- m m 0 3 3
F = (k ) 1 -- x m 0 1 0 0
We can make this system dimensionless by using as the time scale: .  The dimensionless system will be written as
,
where now all the and should be interpreted as the ratio of the dimensional value to and , respectively.  For example, if all and , then
A = (-2 1 1 ) 1 -2 1 1 1 -2
Also,
X _ 0 = (x ) 0 0 0
Let the eigenvectors of be written as and the eigenvalues be .
Construct a matrix with the columns being the eigenvectors .
Let , which means Y is the amplitude of the eigenvectors that
are summed to represent the displacment.  Here is the clever way
to solve the coupled problem:

d^2/dτ^2 P Y = A P Y + X _ 0, <br />
$FormBox[RowBox[{Cell[], P^(-1)}], TraditionalForm]$

where is a diagonal matrix of the eigenvalues.  Let
.   This is really cool: there are uncoupled O.D.E.s for the amplitudes of each of the eigenvectors:

Finding a particular solution of the above equation is rather elementary.

We next present a technique for finding  the homogeneous solution,
and in particular an efficient way to satisfy the initial conditions
and .  The general homogeneous solution is:
$X _ h(t) = Re { P Y _ h(t) }$
where

Let And let be a column vector of the .
$FormBox[RowBox[{RowBox[{X _ h(0), Cell[]}], =, , P C _ r}], TraditionalForm]$

As a matrix problem, this gives

and

where .

I like to turn off warnings about variable names being similar:

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Create a matrix using "Input"-> "Create Table/Matrix/Pallette".
Then check "Matrix".  The matrix is stored (and can also be input)
as a list of lists, with the inner lists being the rows of A.  Comment
one of these A matrices as text, "uncomment" as needed.  Note the
use of one decimal number forces the use of numerical computation, as opposed to
symbolic.

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A = (-2 1 ) (* change this ' Format > Style ' to ' Input ' to use this one *) 1 -2

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First, enjoy the exercise of finding eigenvalues without using the "canned" routine.
Note from the graph we expect negative roots for λ.

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[Graphics:HTMLFiles/index_56.gif]

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Or use the canned routine:

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We will solve the homogeneous equation by proposing . Obviously, . Let's make a list of these "eigenfrequencies":

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Let Mathematica find the eigenvectors:

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${{0.49999999999999994`, -0.7071067811865475`, 0.5`}, {-0.7071067811865477`, -7.577190362262357`*^-17, 0.7071067811865474`}, {0.49999999999999994`, 0.7071067811865476`, 0.49999999999999994`}}$

Store the eigenvectors are columns in :

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Calculate :

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Have a look at :

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Out[13]//MatrixForm=

( 0.5` -0.7071067811865476` 0.5000000000000001` ) -0.7071067811865477` -1.9968293511751778`*^-16 0.7071067811865474` 0.4999999999999999` 0.7071067811865476` 0.5000000000000001`

Have a look at , just to see that it yields the expected diagonal matrix (except for round-off error):

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Out[14]//MatrixForm=

( -3.4142135623730954` 3.0444397003392965`*^-16 2.937239210534792`*^-16 ) 8.651933336434325`*^-17 -2.` 1.8026216370287118`*^-16 9.074772183703672`*^-17 -1.0928757898653885`*^-16 -0.585786437626905`

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Out[15]//MatrixForm=

( -3.4142135623730954` 0 0 ) 0 -2.` 0 0 0 -0.5857864376269049`

In this particular example, we set . In doing so, we have implicitly nondimensionalized displacement in terms of the amplitude of the "jiggle".

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Project the forcing onto the eigenvectors:

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A little "pencil and paper math" yields the amplitudes of the
eigenvectors as caused by the jiggling:

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${(0.5` Sin[t Ω])/(3.414213562373096` - Ω^2), -(0.7071067811865477` Sin[t Ω])/(2.0000000000000004` - Ω^2), (0.4999999999999999` Sin[t Ω])/(0.5857864376269049` - Ω^2)}$

Let's consider a particular value of Ω, so that we can have something
to plot out:

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${{0.9238795325112867`, -1.0000000000000002`, 0.38268343236508967`}, {-1.3065629648763766`, -1.0715765374994131`*^-16, 0.541196100146197`}, {0.9238795325112868`, 0.9999999999999998`, 0.38268343236508967`}}$

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Now with all the masses assumed to have no displacement and no velocity at t=0, we have

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It is amazing what Mathematica can do with a list. Exp[list] makes a list. list*list makes a list of the product of the elements...

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${(0.` + 0.04275913188839186` i) e^(1.8477590650225737` i t), (0.` - 0.14285714285714285` i) e^(1.4142135623730951` i t), (0.` + 0.9727633537779373` i) e^(0.7653668647301795` i t)}$

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Check that we really did satisfy initial conditions:

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${-1.1102230246251565`*^-16, 0.`, -5.551115123125783`*^-17}$

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[Graphics:HTMLFiles/index_118.gif]

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[Graphics:HTMLFiles/index_121.gif]

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[Graphics:HTMLFiles/index_124.gif]

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Check close up to see that initial conditions are really satisfied:

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[Graphics:HTMLFiles/index_127.gif]

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Here we plot all the , with a shift added to the displacements to
make them easier to see. The shift makes a plot look like the
displacement from a common origin, as it might naturally appear
if the masses are hanging from a ceiling.

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[Graphics:HTMLFiles/index_132.gif]

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Converted by Mathematica (March 3, 2003)

Coupled oscillators,jiggled from one end

Coupled oscillators,
jiggled from one end