Question

The equation of motion of a spring-mass system with damping (see Section 3.8) is $$ m \frac{d^{2} u}{d t^{2}}+c \frac{d u}{d t}+k u=0 $$ where \(m, c,\) and \(k\) are positive. Write this second order equation as a system of two first order equations for \(x=u, y=d u / d t .\) Show that \(x=0, y=0\) is a critical point, and analyze the nature and stability of the critical point as a function of the parameters \(m, c,\) and \(k .\) A similar analysis can be applied to the electric circuit equation (see Section 3.8) $$L \frac{d^{2} I}{d t^{2}}+R \frac{d I}{d t}+\frac{1}{C} I=0.$$

Short answer

Answer: The critical point of the given spring-mass system with damping is (x, y) = (0, 0). The stability of the critical point depends on the parameters m, c, and k through their effect on the discriminant Δ.

Step-by-step solution

Writing two first-order equations

Let \(x = u\) and \(y = \frac{du}{dt}\). So, we have \(\frac{dx}{dt} = \frac{du}{dt} = y\) and \(\frac{dy}{dt} = \frac{d^2u}{dt^2}\). Now, substitute these into the given equation: $$m\frac{d^2u}{dt^2} + c\frac{du}{dt} + ku = 0 \Rightarrow m\frac{dy}{dt} + cy + kx = 0$$ The system of two first-order equations is: $$ \begin{cases} \frac{dx}{dt} = y \\ \frac{dy}{dt} = -\frac{c}{m}y - \frac{k}{m}x \end{cases} $$ #Step 2: Find the critical points#

Finding critical points

The critical point is when both \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\). $$ \begin{cases} y = 0 \\ -\frac{c}{m}y - \frac{k}{m}x = 0 \end{cases} $$ Since \(y = 0\), the second equation becomes \(-\frac{k}{m}x = 0\). Since m and k are positive, the critical point occurs when \(x = 0\). Thus, the critical point is \((x, y) = (0, 0)\). #Step 3: Analyze the nature and stability of the critical point#

Nature and stability of the critical point

To analyze the nature and stability of the critical point, we need to examine the eigenvalues of the Jacobian matrix of the linear system. Let \(f(x, y) = y\) and \(g(x, y) = -\frac{c}{m}y - \frac{k}{m}x\). The Jacobian matrix (J) is: $$ J = \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x } & \frac{\partial g}{\partial y} \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -\frac{k}{m} & -\frac{c}{m} \end{bmatrix} $$ The eigenvalues of the Jacobian matrix are the solutions of the characteristic equation: $$ \mathrm{det}(J - \lambda I) = \begin{vmatrix} -\lambda & 1 \\ -\frac{k}{m} & -\frac{c}{m} - \lambda \end{vmatrix} = 0 $$ Solving this, we get \(\lambda^2 + \frac{c}{m}\lambda + \frac{k}{m} = 0\). The nature and stability depend on the sign of the discriminant, \(\Delta = \left(\frac{c}{m}\right)^2 - 4\left(\frac{k}{m}\right)\): 1. If \(\Delta > 0\), the system has two distinct real eigenvalues, which implies that the critical point is a saddle point and is unstable. 2. If \(\Delta = 0\), the system has a repeated real eigenvalue, which implies that the critical point is a degenerate node and the nature cannot be determined from this analysis alone. 3. If \(\Delta < 0\), the system has complex conjugate eigenvalues, which implies that the critical point is a spiral point, and it is stable if the real part of the eigenvalues is negative, and unstable otherwise. The stability of the critical point \((0, 0)\) depends on the parameters m, c, and k through their effect on the discriminant \(\Delta\).