Problem 10 - Differential Equations and Linear Algebra

Question

Determine the motion of the spring-mass system governed by the given initial- value problem. In each case, state whether the motion is underdamped, critically damped, or overdamped, and make a sketch depicting the motion. $$\frac{d^{2} y}{d t^{2}}+5 \frac{d y}{d t}+6 y=0, \quad y(0)=-1, \quad \frac{d y}{d t}(0)=4$$

Step-by-Step Solution

Answer

The motion of the spring-mass system is described by the particular solution \(y(t) = 2e^{-2t} - 3e^{-3t}\). Since the roots of the characteristic equation are distinct and negative, the motion is overdamped. This implies that the system will slowly return to its equilibrium position without oscillating. The sketch of the motion will show the function converging to zero over time without oscillations.

Step 1: Write down the differential equation, its characteristic equation, and initial conditions

The given initial-value problem is: \[\frac{d^{2} y}{d t^{2}} + 5 \frac{d y}{d t} + 6 y = 0\] The initial conditions are: \[y(0) = -1,\] \[\frac{d y}{dt}(0) = 4.\] The characteristic equation for the given problem is: \[r^2 + 5r + 6 = 0\]

Step 2: Solve the characteristic equation

To solve the characteristic equation, \(r^2 + 5r + 6 = 0\), we will find its roots. This equation is a quadratic equation which can be factored as: \[(r + 2)(r + 3) = 0\] So, the roots of the characteristic equation are: \[r_1 = -2,\] \[r_2 = -3\]

Step 3: Get the general solution

As the roots are distinct, the general solution is given by: \[y(t) = C_1e^{-2t} + C_2e^{-3t}\]

Step 4: Use the initial conditions to solve for constants

To find the constants \(C_1\) and \(C_2\), we will use the initial conditions: For \(y(0) = -1\): \[-1 = C_1e^{0} + C_2e^{0}\] \[-1 = C_1 + C_2\] For \(\frac{d y}{dt}(0) = 4\), we first need to get the derivative of \(y(t)\): \[\frac{dy}{dt} = -2C_1e^{-2t} - 3C_2e^{-3t}\] Now, we evaluate the derivative at t=0: \[4 = -2C_1e^{0} - 3C_2e^{0}\] \[4 = -2C_1 - 3C_2\] Solving this system of equations, we get: \[C_1 = 2\] \[C_2 = -3\] So, the particular solution for the initial-value problem is: \[y(t) = 2e^{-2t} - 3e^{-3t}\]

Step 5: Classify the motion and sketch it

As both roots are negative and distinct, the motion is overdamped. This means the system will slowly return to its equilibrium position without oscillating. To sketch the motion, plot the function \(y(t) = 2e^{-2t} - 3e^{-3t}\) on the time-domain. The behavior of the function shows that it is an overdamped motion, which doesn't oscillate around its equilibrium and converges to zero over time.

Key Concepts

Differential Equations

Differential equations are mathematical expressions that relate functions to their rates of change. In the context of a spring-mass system, the differential equation describes the motion of the mass as it responds to forces such as tension and damping. In our example, the differential equation is \[\frac{d^{2} y}{d t^{2}} + 5 \frac{d y}{d t} + 6 y = 0\]. This equation is a second-order linear homogeneous differential equation, which is typical for mechanical systems involving a mass, a spring, and a damper.

Here, the terms can be interpreted as follows:

• The term \(\frac{d^{2} y}{d t^{2}}\) represents the acceleration of the mass.
• The term \(5 \frac{d y}{d t}\) represents the damping force, which is proportional to the velocity.
• The term \(6y\) represents the restoring force exerted by the spring.

Understanding how these terms contribute to the system's dynamics is key to solving the differential equation and predicting the system's motion.

Characteristic Equation

The characteristic equation is derived from the differential equation and plays a crucial role in finding its solutions. It is obtained by assuming a solution of the form \(e^{rt}\) and substituting it into the differential equation. For the given problem, this leads to the characteristic equation:\[r^2 + 5r + 6 = 0\].

This quadratic equation is solved to find the roots, which tell us how the system behaves. By factoring the equation, we determine the roots:

• \(r_1 = -2\)
• \(r_2 = -3\)

The roots are distinct and real, which indicates that we have two exponential solutions that can be combined to form the general solution. These roots also help us classify the nature of the system's damping.

Overdamped Motion

Overdamped motion occurs when the damping in a system is strong enough that it prevents oscillations. This is characterized by real, distinct roots in the characteristic equation, as we see in our example with roots \(r_1 = -2\) and \(r_2 = -3\).

In an overdamped system, the motion of the mass returns to equilibrium slowly without ever crossing the equilibrium position. The response is composed of two decaying exponentials:\[y(t) = C_1e^{-2t} + C_2e^{-3t}\].

Overdamped systems typically stabilize more slowly than critically damped systems, but they do so without oscillation. This behavior is important in applications where overshooting the equilibrium position must be avoided.

Initial Value Problem

An initial value problem is a differential equation coupled with specific values at the start of the problem. These values, known as initial conditions, allow us to determine a unique solution from the general solution.

In the given problem, the initial conditions are:

• \(y(0) = -1\)
• \(\frac{d y}{dt}(0) = 4\)

These conditions describe the initial position and velocity of the mass at time \(t = 0\). By plugging these into the general solution, \(y(t) = C_1e^{-2t} + C_2e^{-3t}\), we can calculate the specific constants \(C_1\) and \(C_2\). This yields:

• \(C_1 = 2\)
• \(C_2 = -3\)

Thus, the particular solution \(y(t) = 2e^{-2t} - 3e^{-3t}\) satisfies both the differential equation and the initial conditions, making it unique to this initial value problem.