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