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 \(\Delta\).

Key concepts

Spring-Mass System

In physics, the spring-mass system is a classic example illustrating Newton's laws of motion. Imagine a spring either suspended or horizontal with a mass attached to one end. This system is governed by Hooke's Law, which connects the force exerted by a spring to the displacement of the mass at its end. The simple equation is given as \[ F = -kx \] where:

• \( F \) is the force exerted by the spring
• \( k \) is the spring constant, depicting the stiffness of the spring
• \( x \) is the displacement from the equilibrium position

The spring-mass system can be used to demonstrate simple harmonic motion when there is no friction or resistance, creating a produce, oscillatory system. However, when there is resistance, or damping, the motion changes, as illustrated in the exercise with the introduction of the damping constant \( c \). This complicates the motion by adding a second term which modifies how the mass oscillates over time.

Damping

Damping in a spring-mass system refers to the effect of reducing the vibrations and eventually stopping the motion entirely. This can be thought of as the 'brakes' of the system, slowing it down naturally. Damping is introduced into the model using a term proportional to the velocity. For a damped spring-mass system, the motion equation is given as:\[ m \frac{d^{2} u}{d t^{2}} + c \frac{d u}{d t} + k u = 0 \]where:

• \( m \) is the mass
• \( c \) is the damping coefficient
• \( k \) is the spring constant

The damping term \( c \frac{d u}{d t} \) acts against the velocity, reducing system energy over time. The magnitude of \( c \) determines the type of damping:

• Under-damped: \( c \) is small, system oscillates heavily then calms down
• Critically damped: just right \( c \), returns to equilibrium without oscillating
• Over-damped: large \( c \), system returns to equilibrium very slowly without oscillating

Knowing how damping works allows us to predict how different materials will react to forces, ensuring safety and efficiency in design.

Critical Points

In dynamical systems such as the damped spring-mass model, critical points represent states where the system is in equilibrium – not moving. To find critical points, we set the first derivatives equal to zero: \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \).This leads us to the equations:

• \( y = 0 \)
• \( -\frac{k}{m}x = 0 \)

Here, these conditions show that both position and velocity are zero, indicating the point \( (x, y) = (0, 0) \) as a critical point. It tells us that the system is not moving because the mass is at rest in its equilibrium position with no velocity. This is crucial as it allows us to investigate further into how the system behaves around this state.

Stability Analysis

Understanding stability is essential as it informs us how a system responds to small disturbances. For the spring-mass system, the nature of the critical point (0,0) in terms of stability is analyzed through its associated linearized system. The Jacobian matrix of the system at the critical point helps us find the eigenvalues, which let us assess the system's response.The characteristic equation given by:\[ \lambda^2 + \frac{c}{m}\lambda + \frac{k}{m} = 0 \] allows us to calculate the eigenvalues. The discriminant:\[ \Delta = \left(\frac{c}{m}\right)^2 - 4\left(\frac{k}{m}\right) \] determines their nature:

• If \( \Delta > 0 \), distinct real eigenvalues indicate an unstable saddle point.
• If \( \Delta = 0 \), repeated eigenvalues result in a degenerate node whose stability depends on higher-order terms.
• If \( \Delta < 0 \), complex conjugate eigenvalues point to a spiral which is stable if the real parts are negative.

This analysis provides a framework to understand how changes in \( m, c, \) or \( k \) can lead to different stability outcomes for the system's equilibrium state.