10.6: The Mass-Spring-Damper System
( \newcommand{\kernel}{\mathrm{null}\,}\)
Figure 1. Mass, spring, damper system
If one provides an initial displacement, x0, and velocity, v0, to the mass depicted in Figure then one finds that its displacement, x(t) at time t satisfies
md2x(t)dt2+2cdx(t)dt+kx(t)=0
x(0)=x0
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
u1(t)=x(t)
u2(t)=x′(t)
and note that Equation becomes
mu2′(t)=(−(ku1(t)))−2cu2(t)
u1′(t)=u2(t)
Denoting u(t)≡(u1(t)u2(t))T we write Equation as
∀A,A=(01−km−2cm):(u′(t)=Au(t))
We recall from The Matrix Exponential module that
u(t)=eAtu(0)
We shall proceed to compute the matrix exponential along the lines of The matrix Exponential via Eigenvalues and Eigenvectors module. To begin we record the resolvent
R(z)=−1mz2+2cz+k(2c+mzm−kmz)
The eigenvalues are the roots of mz2+2cz+k
∀d,d=√c2−mk:(λ1=(−c)−dm
∀d,d=√c2−mk:(λ2=(−c)+dm
We naturally consider two cases, the first being
- d≠0. In this case the partial fraction expansion of R(z) yields
R(z)=−1z−λ112d(d−c−mkc+d)+−1z−λ212d(c+dm−kd−c)=−1z−λ1P1+−1z−λ2P2
and so eAt=eλ1tP1+eλ2tP2 i.e., v0 it follows that
x(t)=x02d(eλ1t(d−c)+eλ2t(c+d))
If d is real, i.e., if c2>mk then both λ1 and λ2 are negative real numbers and x(t) decays to 0 without oscillation. If, on the contrary, d is imaginary, i.e., c2<mk, then
x(t)=e−(ct)(cos(|d|t)+c|d|sin(|d|t))
and so x decays to 0 in an oscillatory fashion. When Equation holds the system is said to be overdamped while when Equation governs then we speak of the system as underdamped. It remains to discuss the case of critical damping.
- d=0. In this case, λ1=λ2=−√km 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 P1=I. Similarly, this equation dictates that
D1=AP1−λ1P1=A−λ1I=(√km1−km√km)
On substitution of this into this equation we find
eAt=e−(t√km)(1+t√km1−(tkm)1−t√km)
Under the assumption, as above, that v0=0, we deduce from Equation that
x(t)=e−(t√km)(1+t√km)x0