Question

The equation of motion of a mass-spring system with damping is given by $$ m \ddot{x}+c \dot{x}+k x=0 $$ where \(m, c\) and \(k\) are positive constants. By changing this equation into a system, discuss the nature and stability of the critical point.

Short answer

Answer: In order to determine the stability of the system, we can perform the following steps: 1. Introduce a new variable to represent the first derivative term (velocity). 2. Convert the second-order ODE into a system of two first-order ODEs. 3. Find the critical (steady state) point of the system by setting the time derivatives equal to zero. 4. Linearize the system and find the Jacobian matrix. 5. Analyze the stability using the eigenvalues of the Jacobian matrix. The stability of the mass-spring system with damping will depend on the eigenvalues of the Jacobian matrix of the linearized system. The nature and stability of the critical point can be determined by analyzing the real part of the eigenvalues.

Step-by-step solution

Introduce a new variable

Let's introduce a new variable \(v\) such that \(v = \dot{x}\). That is, the velocity is the derivative of the displacement with respect to time. The given equation now becomes: $$ m \dot{v} + c v + kx = 0 $$

Convert the second-order ODE into a system of first-order ODEs

Now we have two first-order ODEs for the displacement \(x\) and the velocity \(v\): $$ \begin{cases} \dot{x} = v \\ \dot{v} = -\frac{c}{m} v - \frac{k}{m} x \end{cases} $$

Find the critical point (steady state) of the system

A critical point (steady state) occurs when the time derivatives are equal to zero. That is, \(\dot{x}=0\) and \(\dot{v}=0\). From our system of ODEs, we have: $$ \begin{cases} 0 = v \\ 0 = -\frac{c}{m} v - \frac{k}{m} x \end{cases} $$ Since \(v=0\), the second equation simplifies to \(0 = -\frac{k}{m} x\). Thus, the critical point (steady state) is \((x, v) = (0, 0)\).

Linearize the system and find the Jacobian matrix

Let's linearize the system by expressing the system as \(\dot{\mathbf{x}} = A\mathbf{x}\), where \(\mathbf{x} = \begin{bmatrix}x\\ v\end{bmatrix}\) and \(A\) is the Jacobian matrix given by: $$ A = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial v} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial v} \end{bmatrix}_{\mathbf{x}=(0,0)} = \begin{bmatrix} 0 & 1 \\ -\frac{k}{m} & -\frac{c}{m} \end{bmatrix} $$

Analyze the stability using eigenvalues of the Jacobian matrix

The stability of the critical point is determined by the eigenvalues of the Jacobian matrix \(A\). The eigenvalues can be found by solving the characteristic equation of the matrix: $$ \text{det}(A - \lambda I) = \begin{vmatrix} -\lambda & 1 \\ -\frac{k}{m} & -\frac{c}{m} - \lambda \end{vmatrix} = \lambda^2 + \frac{c}{m} \lambda + \frac{k}{m} = 0 $$ The sign of the real part of the eigenvalues determines the stability of the system at the critical point: - If the real part of both eigenvalues is negative, the system is stable and the critical point is a stable node (or a stable spiral if the eigenvalues are complex with a negative real part). - If the real part of both eigenvalues is positive, the system is unstable and the critical point is an unstable node (or unstable spiral if the eigenvalues are complex with a positive real part). - If the real part of one eigenvalue is positive and the other is negative, the system is a saddle point and the critical point is unstable. In summary, the stability of the mass-spring system with damping depends on the eigenvalues of the Jacobian matrix of the linearized system. The nature and stability of the critical point can be determined by analyzing the real part of the eigenvalues.

Key concepts

Mass-Spring System

The mass-spring system is a classic example in physics and engineering. Imagine a simple mechanical system with a mass attached to a spring. As the mass moves, it compresses or stretches the spring. This movement creates a force that acts to restore the mass to its original position, leading to oscillations. The force is described by Hooke's Law, where the force is proportional to the displacement.

In mathematical terms, the motion of a mass-spring system is typically expressed as a second-order differential equation: \[ m \ddot{x} + c \dot{x} + k x = 0 \]Here, \(m\) represents the mass, \(c\) is the damping coefficient, and \(k\) is the spring constant. The term \(c \dot{x}\) accounts for damping, which resists the motion and gradually diminishes the oscillations.

Damping plays a crucial role in real-world applications, where systems need stability. Without damping, the oscillations would continue indefinitely, like an ideal pendulum. With damping, the system will eventually come to rest in a stable state, making it easier to analyze and predict its behavior.

Stability Analysis

Stability analysis in differential equations helps us understand how solutions behave as time progresses. For a mass-spring system, we conduct a stability analysis to determine the behavior around a critical point (steady state).

A critical point occurs when the rate of change is zero, here denoted as \((x, v) = (0, 0)\). To perform stability analysis, we linearize the system of equations and find the critical point. We express it as a system of first-order ODEs:

• \( \dot{x} = v \)
• \( \dot{v} = -\frac{c}{m} v - \frac{k}{m} x \)

With these equations, we can analyze how the system behaves near the steady state. This is crucial because knowing whether a system returns to equilibrium or diverges can inform us about the system's reliability in practical applications.

Ultimately, the behavior of the solutions at the critical point tells us whether it's stable (returns to equilibrium) or unstable (moves away from equilibrium). Stability analysis serves as a backbone in understanding dynamic systems, which has a multitude of applications in fields ranging from mechanical engineering to biological systems.

Jacobian Matrix

The Jacobian matrix is a mathematical tool that provides critical insights into the local behavior of a system of differential equations. For the mass-spring system, after converting its differential equation into a system of first-order equations, we derive the Jacobian matrix to analyze stability.

For our system of equations:

• \( \dot{x} = v \)
• \( \dot{v} = -\frac{c}{m} v - \frac{k}{m} x \)

We construct the Jacobian at the critical point \((x, v) = (0, 0)\):\[A = \begin{bmatrix}0 & 1 \-\frac{k}{m} & -\frac{c}{m}\end{bmatrix}\]The entries in this matrix represent the partial derivatives of each equation in the system with respect to the variables \(x\) and \(v\).

The eigenvalues of the Jacobian matrix help us determine the system’s stability. If all eigenvalues have negative real parts, the system gravitates back to a state of balance. Conversely, if any eigenvalue possesses a positive real part, disturbances grow and the system can be deemed unstable.

Understanding the Jacobian is essential as it assists in predicting whether a system remains stable under small perturbations around equilibrium. Its application stretches across a variety of scientific disciplines, making it a fundamental concept in the study of dynamical systems.