Math 2250-4
Wed Dec 4
7.4 Mass-spring systems: untethered mass-spring trains, and forced oscillation non-homogeneous
problems.
In your homework for this week you model the spring system below, with no damping. Although we
draw the picture horizontally, it would also hold in vertical configuration if we measure displacements
from equilibrium in the underlying gravitational field.
Let's make sure we understand why the natural system of DEs and IVP for this system is
m1 x1 ## t
m2 x2 ## t
x1 0
x2 0
=Kk1 x1 C k2
=Kk2 x2 K x1
= a 1, x1 # 0
= b 1, x2 # 0
x2 K x1
K k3 x2
= a2
= b2
Exercise 1a) What is the dimension of the solution space to this homogeneous linear system of differential
equations? Why?
1b) What if one had a configuration of n masses in series, rather than just 2 masses? What would the
dimension of the homogeneous solution space be in this case? Why? Examples:
We can write the system of DEs for the system at the top of the page in matrix-vector form:
m1 0
x1 ## t
0 m2
x2 ## t
=
Kk1 K k2
k2
x1
k2
Kk2 K k3
x2
.
We denote the diagonal matrix on the left as the "mass matrix" M, and the matrix on the right as the spring
constant matrix K (although to be completely in sync with Chapter 5 it would be better to call the spring
matrix KK). All of these configurations of masses in series with springs can be written as
M x## t = K x .
If we divide each equation by the reciprocal of the corresponding mass, we can solve for the vector of
accelerations:
x1 ## t
x2 ## t
k1 C k2
k2
m1
m1
K
=
k2
m2
x1
K
k2 C k3
x2
,
m2
which we write as
x## t = A x .
(You can think of A as the "acceleration" matrix.)
Notice that the simplification above is mathematically identical to the algebraic operation of multiplying the
first matrix equation by the (diagonal) inverse of the diagonal mass matrix M. In all cases:
M x## t = K x
0 x## t = A x , with A = M K1 K .
How to find a basis for the solution space to conserved-energy mass-spring systems of DEs
x## t = A x .
Based on our previous experiences, the natural thing to for this homogeneous system of linear differential
equations would be to try and find a basis of solutions of the form
x t = er t v
(We would maybe also think about first converting the second order system to an equivalent first order
system of twice as many DE's, one for for each position function and one for each velocity function.) But
there's a shortcut we'll take instead, and here's why: For a single mass-spring configuration with no
damping the corresponding scalar equation was
2
2
x## t =Kw0 x i.e. x## t C w0 x = 0.
The roots of the characteristic polynomial were purely imaginary and after using Euler's formula we
realized the real-value solution functions were
x t = A cos w0 t C B sin w0 t = C cos w0 t K a .
Now, in the present case of systems of masses and springs there is also no damping. Thus, the total
energy - consisting of the sum of kinetic and potential energy - will always be conserved. Thus any
solution of the form
x t = er t v = e a C w i t v
cannot grow or decay exponentially, so must have a = 0: If a O 0 the total energy would increase
exponentially, and if a ! 0 it would decay to zero exponentially. In other words, in order for the total
energy to remain constant we must actually have
x t = ei w t v = cos w t C i sin w t v .
So, we skip the exponential solutions altogether, and go directly to finding homogeneous solutions of the
form
cos w t v sin w t v.
To find what w, v must be, we substitute x t = cos w t v (or x t = sin w t v) into the DE system:
2
x## t = A x
x t = cos w t v
0 Kw cos w t v = A cos w t v = cos w t A v .
This identity will hold c t if and only if
2
Kw v = A v.
2
So, v must be an eigenvector of A, but the corresponding eigenvalue is l =Kw . In other words, the
angular frequency w is given by w = Kl where l is the eigenvalue with eigenvector v. If we used a trial
solution y t = sin w t v we would arrive at the same eigenvector equation. This leads to the
Solution space algorithm: Consider a very special case of a homogeneous system of linear differential
equations,
x## t = A x .
If An # n is a diagonalizable matrix and if all of its eigenvalues are negative, then for each eigenpair
l j , vj there are two linearly independent solutions to x## t = A x given by
xj t = cos wj t vj
yj t = sin wj t vj
with
wj = Kl j .
This procedure constructs 2 n independent solutions to the system x## t = A x, i.e. a basis for the
solution space.
Remark: What's amazing is that the fact that if the system is conservative the acceleration matrix will
always be diagonalizable, and all of its eigenvalues will be non-positive. In fact, if the system is tethered to
at least one wall (as in the first two diagrams on page 1), all of the eigenvalues will be strictly negative, and
the algorithm above will always yield a basis for the solution space. (If the system is not tethered and is
free to move as a train, like the third diagram on page 1, then l = 0 will be one of the eigenvalues, and will
yield the constant velocity and displacement contribution to the solution space, c1 C c2 t v, where v is the
corresponding eigenvector. Together with the solutions from strictly negative eigenvalues this will still
lead to the general homogeneous solution.)
Exercise 2) Consider the special case of the configuration on page one for which m1 = m2 = m and
k1 = k2 = k3 = k . In this case, the equation for the vector of the two mass accelerations reduces to
2k
Km
x ## t
1
x ## t
=
k
2
m
k
= m
K2
1
1
K2
K2
1
1
K2
k
m
x
2k
Km
x
1
2
x
1
x
.
2
a) Find the eigendata for the matrix
.
k
times this matrix.
m
c) Find the 4Kdimensional solution space to this two-mass, three-spring system.
b) Deduce the eigendata for the acceleration matrix A which is
solution The general solution is a superposition of two "fundamental modes". In the slower mode both
k
masses oscillate "in phase", with equal amplitudes, and with angular frequency w1 =
. In the faster
m
mode, both masses oscillate "out of phase" with equal amplitudes, and with angular frequency
3k
w2 =
. The general solution can be written as
m
x1 t
1
1
= C1 cos w1 t K a 1
C C2 cos w2 t K a 2
x2 t
1
K1
1
= c1 cos w1 t C c2 sin w1 t
1
C c3 cos w2 t C c4 sin w2 t
1
K1
.
Exercise 3) Show that the general solution above lets you uniquely solve each IVP uniquely. This should
reinforce the idea that the solution space to these two second order linear homgeneous DE's is four
dimensional.
2k
Km
x ## t
1
x ## t
2
=
k
m
k
m
x
2k
Km
x
1
2
x1 0 = a1 , x1 # 0 = a2
x2 0 = b1 , x2 # 0 = b2
Experiment: Although we won't have time to predict the numerical values for the two fundamental modes
of the vertical mass-spring configuration corresponding to Exercise 2, and then check our predictions like
we did for the single mass-spring configuration, I have brought along a demonstration so that we can see
these two vibrations.
Forced oscillations (still undamped):
M x## t = K x C F t
0 x## t = A x C M K1 F t .
If the forcing is sinusoidal,
M x## t = K x C cos w t G0
0 x## t = A x C cos w t F0
with F0 = M K1 G0 .
From the fundamental theorem for linear transformations we know that the general solution to this
inhomogeneous linear problem is of the form
x t = xP t C xH t ,
and we've been discussing how to find the homogeneous solutions xH t .
As long as the driving frequency w is NOT one of the natural frequencies, we don't expect resonance; the
method of undetermined coefficients predicts there should be a particular solution of the form
xP t = cos w t c
where the vector c is what we need to find.
Exercise 4) Substitute the guess xP t = cos w t c into the DE system
x## t = A x C cos w t F0
to find a matrix algebra formula for c = c w . Notice that this formula makes sense precisely when w is
NOT one of the natural frequencies of the system.
Solution:
2
c w =K A C w I
2
K1
F0 .
Note, matrix inverse exists precisely if Kw is not an eigenvalue.
Exercise 5) Continuing with the configuration from page 1, but now for an inhomogeneous forced
problem, let k = m , and force the second mass sinusoidally:
x ## t
1
x ## t
=
K2
1
1
K2
2
x
1
C cos w t
x
2
0
3
We know from previous work that the natural frequencies are w1 = 1, w2 =
xH t = C1 cos t K a 1
1
3 t K a2
C C2 cos
3 and that
1
.
1
K1
Find the formula for xP t , as on the preceding page. Notice that this steady periodic solution blows up
as w/1 or w/ 3 . (If we don't have time to work this by hand, we may skip directly to the
technology check on the next page. But since we have quick formulas for inverses of 2 by 2 matrices, this
is definitely a computation we could do by hand.)
Solution: As long as w s 1,
3 , the general solution x = xP C xH is given by
3
x t
1
x t
2
= cos w t
2
w K1
w K3
6K3 w
2
2
2
w K1
C C1 cos t K a 1
2
w K3
1
C C2 cos
3 t K a2
1
.
1
K1
Interpretation as far as inferred practical resonance for slightly damped problems: If there was even a
small amount of damping, the homogeneous solution would actually be transient (it would be
exponentially decaying and oscillating - underdamped). There would still be a sinusoidal particular
solution, which would have a formula close to our particular solution, the first term above, as long as
w s 1, 3 . (There would also be a relatively smaller sin w t d term as well.) So we can infer the
practical resonance behavior for different w values with slight damping, by looking at the size of the c w
term for the undamped problem....see next page for visualizations.
>
>
>
>
>
restart :
with LinearAlgebra :
A d Matrix 2, 2, K2, 1, 1,K2
F0 d Vector 0, 3 :
Iden d IdentityMatrix 2 :
2
> c d w/ A C w $Iden
> cw ;
:
K1
. KF0 : # the formula we worked out by hand
3
2
3K4 w Cw
K
3 K2 C w
4
(1)
2
2
3K4 w Cw
4
> with plots :
with LinearAlgebra :
> plot Norm c w , 2 , w = 0 ..4, magnitude = 0 ..10, color = black, title = `practical resonance` ;
# Norm(c(w),2) is the magnitude of the c(w) vector
practical resonance
10
8
magnitude
6
4
2
0
> plot Norm c
0
1
2
w
3
4
2$Pi
, 2 , T = 0 ..15, magnitude = 0 ..15, color = black, title
T
= `practical resonance as function of forcing period` ;
practical resonance as function of forcing period
15
10
magnitude
5
0
0
5
10
T
>
15