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The Mass-Spring-Damper System

Module by: Steven J. Cox. E-mail the author

Summary: This module gives an complicate example of solving matrix exponential problem.

Figure 1: Mass, spring, damper system
Figure 1 (msd.png)

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

md2xtdt2+2cdxtdt+kxt=0 m t2 x t 2 c t x t k x t 0
(1)
x0= x0 x 0 x0 x0= 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 u1 t=xt u1 t x t u2 t=xt u2 t x t and note that Equation 1 becomes
m u2 t=((k u1 t))2c u2 t m u2 t k u1 t 2 c u2 t
(2)
u1 t= u2 t u1 t u2 t Denoting ut( u1 t u2 t )T u t u1 t u2 t we write Equation 2 as
ut=Aut  ,   A=( 01 km-2cm )    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+k( 2c+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 λ1 =(c)dm  ,   d=c2mk    d d c 2 m k λ1 c d m λ2 =c+dm  ,   d=c2mk    d d c 2 m k λ2 c d m We naturally consider two cases, the first being

  1. d0d0. In this case the partial fraction expansion of Rz R z yields Rz=-1z λ1 12d( dcm kc+d )+-1z λ2 12d( c+dm kdc )=-1z λ1 P1 +-1z λ2 P2 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=e λ1 t P1 +e λ2 t P2 A t λ1 t P1 λ2 t P2 . If we now suppose a negligible initial velocity, i.e., v0 v0 , it follows that
    xt= x0 2d(e λ1 t(dc)+e λ2 t(c+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(ct)(cos|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.
  2. 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 =A P1 λ1 P1 =A λ1 I=( km1 kmkm ) 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(tkm)( 1+tkmt (tkm)1tkm ) 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(tkm)(1+tkm) x0 x t t k m 1 t k m x0

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