10.6: The Mass-Spring-Damper System

Figure 1. Mass, spring, damper system

If one provides an initial displacement, \(x_{0}\), and velocity, \(v_{0}\), to the mass depicted in Figure then one finds that its displacement, \(x(t)\) at time \(t\) satisfies

\[m \frac{d^{2}x(t)}{dt^2}+2c \frac{dx(t)}{dt}+kx(t) = 0 \nonumber\]

\[x(0) = x_{0} \nonumber\]

\[x′(0) = v_{0} \nonumber\]

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

\[u_{1}(t) = x(t) \nonumber\]

\[u_{2}(t) = x′(t) \nonumber\]

and note that Equation becomes

\[mu_{2}′(t) = (-(ku_{1}(t)))-2cu_{2}(t) \nonumber\]

\[u_{1}′(t) = u_{2}(t) \nonumber\]

Denoting \(u(t) \equiv \begin{pmatrix} {u_{1}(t)}&{u_{2}(t)} \end{pmatrix}^T\) we write Equation as

\[\forall A, A = \begin{pmatrix} {0}&{1}\\ {\frac{-k}{m}}&{\frac{-2c}{m}} \end{pmatrix} : (u′(t) = Au(t)) \nonumber\]

We recall from The Matrix Exponential module that

\[u(t) = e^{At}u(0) \nonumber\]

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

\[R(z) = \frac{-1}{mz^2+2cz+k} \begin{pmatrix} {2c+mz}&{m}\\ {-k}&{mz} \end{pmatrix} \nonumber\]

The eigenvalues are the roots of \(mz^2+2cz+k\)

\[\forall d, d = \sqrt{c^2-mk} : (\lambda_{1} = \frac{(-c)-d}{m} \nonumber\]

\[\forall d, d = \sqrt{c^2-mk} : (\lambda_{2} = \frac{(-c)+d}{m} \nonumber\]

We naturally consider two cases, the first being

• \(d \ne 0\). In this case the partial fraction expansion of \(R(z)\) yields

\[R(z) = \frac{-1}{z-\lambda_{1}} \frac{1}{2d} \begin{pmatrix} {d-c}&{-m}\\ {k}&{c+d} \end{pmatrix}+\frac{-1}{z-\lambda_{2}} \frac{1}{2d} \begin{pmatrix} {c+d}&{m}\\ {-k}&{d-c} \end{pmatrix} = \frac{-1}{z-\lambda_{1}} P_{1}+\frac{-1}{z-\lambda_{2}} P_{2} \nonumber\]

and so \(e^{At} = e^{\lambda_{1}t}P_{1}+e^{\lambda_{2}t}P_{2}\) i.e., \(v_{0}\) it follows that

\[x(t) = \frac{x_{0}}{2d} (e^{\lambda_{1}t}(d-c)+e^{\lambda_{2}t}(c+d)) \nonumber\]

If \(d\) is real, i.e., if \(c^2 > mk\) then both \(\lambda_{1}\) and \(\lambda_{2}\) are negative real numbers and \(x(t)\) decays to 0 without oscillation. If, on the contrary, \(d\) is imaginary, i.e., \(c^2 < mk\), then

\[x(t) = e^{-(ct)}(\cos(|d|t)+\frac{c}{|d|}\sin(|d|t)) \nonumber\]

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, \(\lambda_{1} = \lambda_{2} = -\sqrt{\frac{k}{m}}\) and so we need only compute \(P_{1}\) and \(D_{1}\). As there is but one \(P_{j}\) and the \(P_{j}\) are known to sum to the identity it follows that \(P_{1} = I\). Similarly, this equation dictates that

\[D_{1} = AP_{1}-\lambda_{1}P_{1} = A-\lambda_{1}I = \begin{pmatrix} {\sqrt{\frac{k}{m}}}&{1}\\ {-\frac{k}{m}}&{\sqrt{\frac{k}{m}}} \end{pmatrix} \nonumber\]

On substitution of this into this equation we find

\[e^{At} = e^{-(t\sqrt{\frac{k}{m}})} \begin{pmatrix} {1+t\sqrt{\frac{k}{m}}}&{1}\\ {-(t\frac{k}{m})}&{1-t\sqrt{\frac{k}{m}}} \end{pmatrix} \nonumber\]

Under the assumption, as above, that \(v_{0} = 0\), we deduce from Equation that

\[x(t) = e^{-(t\sqrt{\frac{k}{m}})} (1+t \sqrt{\frac{k}{m}})x_{0} \nonumber\]