The Mass-Spring-Damper System2.32002/06/272002/07/10 00:00:00.003 GMT-5StevenCoxcox@rice.eduLiqunWangliqun@rice.eduStevenCoxcox@rice.edumassspringdamperThis module gives an complicate example of solving matrix exponential problem.
Mass, spring, damper system
If one provides an initial displacement,
x0,
and velocity,
v0,
to the mass depicted in
then one finds that its displacement,
xt at time t satisfies
mt2xt2ctxtkxt0x0x0x0v0
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
u1txtu2txt
and note that becomes
mu2tku1t2cu2tu1tu2t
Denoting
utu1tu2t we write as
AA01km-2cmutAut
We recall from The Matrix
Exponential module that
utAtu0
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-1mz22czk2cmzmkmz
The eigenvalues are the roots of
mz22czk, namely
ddc2mkλ1cdmddc2mkλ2cdm
We naturally consider two cases, the first being d0.
In this case the partial fraction expansion of
Rz yields
Rz-1zλ112ddcmkcd-1zλ212dcdmkdc-1zλ1P1-1zλ2P2
and so
Atλ1tP1λ2tP2. If we now suppose a negligible initial velocity,
i.e.,
v0, it follows that
xtx02dλ1tdcλ2tcd
If d is real, i.e., if
c2mk, then both
λ1
and
λ2
are negative real numbers and
xt decays to 0 without oscillation. If, on the contrary,
d is imaginary,
i.e.,
c2mk, then
xtctdtcddt
and so x decays to 0 in an
oscillatory fashion. When
holds the system is said to be overdamped
while when governs then we
speak of the system as underdamped. It
remains to discuss the case of critical
damping.
d0. In this case
λ1λ2km, and so we need only compute
P1
and
D1.
As there is but one
Pj
and the
Pj
are known to sum to the identity it follows that
P1I. Similarly, this equation dictates that
D1AP1λ1P1Aλ1Ikm1kmkm
On substitution of this into this equation we find
Attkm1tkmttkm1tkm
Under the assumption, as above, that
v00, we deduce from that
xttkm1tkmx0