If one provides an initial displacement,
x0x0,
and velocity,
v0v0,
to the mass depicted in Figure 1
then one finds that its displacement,
xt
x
t
at time tt satisfies
md2dt2xt+2cddtxt+kxt=0
m
t2
x
t
2
c
t
x
t
k
x
t
0
(1)
x0=x0
x
0
x0
x
′0=v0
x
0
v0
where prime denotes differentiation with respect to time. It is
customary to write this single second order equation as a pair
of first order equations. More precisely, we set
u1t=xt
u1
t
x
t
u2t=
x
′t
u2
t
x
t
and note that
Equation 1 becomes
m
u2
′t=-ku1t-2cu2t
m
u2
t
k
u1
t
2
c
u2
t
(2)
u1
′t=u2t
u1
t
u2
t
Denoting
ut≡u1tu2tT
u
t
u1
t
u2
t
we write
Equation 2 as
∀A,A=01-km-2cm:
u
′t=Aut
A
A
0 1
k
m
-2
c
m
u
t
A
u
t
(3)
We recall from
The Matrix
Exponential module that
ut=ⅇAtu0
u
t
A
t
u
0
We shall proceed to compute the matrix exponential along the
lines of
The Matrix
Exponential via Eigenvalues and Eignevectors module.
To begin we record the resolvent
Rz=-1mz2+2cz+k2c+mzm-kmz
R
z
-1
m
z
2
2
c
z
k
2
c
m
z
m
k
m
z
The eigenvalues are the roots of
mz2+2cz+k
m
z
2
2
c
z
k
, namely
∀d,d=c2-mk:λ1=-c-dm
d
d
c
2
m
k
λ1
c
d
m
∀d,d=c2-mk:λ2=-c+dm
d
d
c
2
m
k
λ2
c
d
m
We naturally consider two cases, the first being
-
d≠0d0.
In this case the partial fraction expansion of
Rz
R
z
yields
Rz=-1z-λ112dd-c-mkc+d+-1z-λ212dc+dm-kd-c=-1z-λ1P1+-1z-λ2P2
R
z
-1
z
λ1
1
2
d
d
c
m
k
c
d
-1
z
λ2
1
2
d
c
d
m
k
d
c
-1
z
λ1
P1
-1
z
λ2
P2
and so
ⅇAt=ⅇλ1tP1+ⅇλ2tP2
A
t
λ1
t
P1
λ2
t
P2
. If we now suppose a negligible initial velocity,
i.e.,
=v0
v0
, it follows that
xt=x02dⅇλ1td-c+ⅇλ2tc+d
x
t
x0
2
d
λ1
t
d
c
λ2
t
c
d
(4)
If dd is real, i.e., if
c2>mk
c
2
m
k
, then both
λ1λ1
and
λ2λ2
are negative real numbers and
xt
x
t
decays to 0 without oscillation. If, on the contrary,
dd is imaginary,
i.e.,
c2<mk
c
2
m
k
, then
xt=ⅇ-ctcos|d|t+c|d|sin|d|t
x
t
c
t
d
t
c
d
d
t
(5)
and so xx decays to 0 in an
oscillatory fashion. When Equation 4
holds the system is said to be overdamped
while when Equation 5 governs then we
speak of the system as underdamped. It
remains to discuss the case of critical
damping.
-
d=0d0. In this case
λ1=λ2=-km
λ1
λ2
k
m
, and so we need only compute
P1P1
and
D1D1.
As there is but one
PjPj
and the
PjPj
are known to sum to the identity it follows that
P1=I
P1
I
. Similarly, this equation dictates that
D1=AP1-λ1P1=A-λ1I=km1-km-km
D1
A
P1
λ1
P1
A
λ1
I
k
m
1
k
m
k
m
On substitution of this into this equation we find
ⅇAt=ⅇ-tkm1+tkmt-tkm1-tkm
A
t
t
k
m
1
t
k
m
t
t
k
m
1
t
k
m
(6)
Under the assumption, as above, that
v0=0
v0
0
, we deduce from Equation 6 that
xt=ⅇ-tkm1+tkmx0
x
t
t
k
m
1
t
k
m
x0
