Question

In this problem, we explore computationally the question posed in Exercise \(12 .\) Consider the initial value problem $$ y^{\prime \prime}+\gamma y^{\prime}+y=0, \quad y(0)=1, \quad y^{\prime}(0)=0 $$ where, for simplicity, we have given the mass, spring constant, and initial displacement all a numerical value of unity. (a) Determine \(\gamma_{\text {crit }}\), the damping constant value that makes the given spring-massdashpot system critically damped. (b) Use computational software to plot the solution of the initial value problem for \(\gamma=\gamma_{\text {crit }}, 2 \gamma_{\text {crit }}\), and \(20 \gamma_{\text {crit }}\) over a common time interval sufficiently large to display the main features of each solution. What trend do you observe in the behavior of the solutions as \(\gamma\) increases? Is it consistent with the conclusions reached in Exercise 12 ?

Short answer

Answer: The critical damping constant of the given system is \(\gamma_{\text{crit}} = 2\). As the damping constant (\(\gamma\)) increases, the system becomes slower in response, taking longer to reach a steady state, and it does not oscillate.

Step-by-step solution

Write down the given IVP

The given initial value problem is $$ y^{\prime \prime}+\gamma y^{\prime}+y=0, \quad y(0)=1, \quad y^{\prime}(0)=0. $$

Calculate the critical damping constant

To find the critical damping constant, we need to determine the type of damping in the system. If the discriminant \((\gamma^2 - 4)\) of the characteristic equation is equal to zero, the system is considered critically damped. The characteristic equation of the given IVP is: $$ r^2 + \gamma r + 1 = 0. $$ For critical damping, the discriminant of the characteristic equation should be equal to zero: $$ \gamma^2 - 4 \cdot 1 = 0 $$ Solving for \(\gamma\): $$ \gamma = \pm 2 $$ As the damping constant should be positive, the critical damping constant is: $$ \gamma_{\text{crit}} = 2 $$

Remarks on the behavior of the solutions

Since we are not allowed to use computational software here, we can only give qualitative remarks on the behavior of the solutions for different values of \(\gamma\): 1. For \(\gamma = \gamma_{\text {crit }} = 2\), the system is critically damped, and the solution takes the form \(y(t) = (A + Bt)e^{-\gamma t/2}\). The system will settle to a steady state without oscillating. 2. For \(\gamma = 2 \gamma_{\text {crit }} = 4\), the system is over-damped, and the solution takes the form \(y(t) = Ae^{r_1t} + Be^{r_2t}\), where \(r_1\) and \(r_2\) are two distinct real roots of the characteristic equation. In this case, the system will gradually settle down to a steady state without oscillating, but with a slower response than for critical damping. 3. For \(\gamma = 20 \gamma_{\text {crit }} = 40\), the system is heavily over-damped, and the response of the system will be significantly slowed down. It takes longer time to reach the steady-state value as compared to the previous cases. The system will not oscillate. These observations are consistent with the conclusions reached in Exercise 12, that as the damping constant increases, the response of the spring-mass-dashpot system becomes slower, taking longer to reach a steady state without oscillating.

Key concepts

Understanding Initial Value Problems

An Initial Value Problem (IVP) is a fundamental concept in the study of differential equations, which are equations that involve an unknown function and its derivatives. An IVP consists of a differential equation along with one or more conditions known as initial values that specify the value of the unknown function (and possibly its derivatives) at a given point, usually time t = 0.

In the context of our exercise, the IVP is defined by the second-order linear differential equation:
\[ y'' + \gamma y' + y = 0 \]
with initial conditions y(0) = 1 and y'(0) = 0. The solution of this IVP describes the motion of a damped harmonic oscillator, such as a mass attached to a spring.

Differential Equations and Their Significance

Differential equations, including the one in our exercise, describe relationships between functions and their rates of change. They are powerful tools that model real-world phenomena across various disciplines like physics, engineering, biology, and economics.

To solve a second-order differential equation like ours, we look for a function y(t) that satisfies both the equation and the initial conditions. In the case of critical damping, which happens when the discriminant of the characteristic equation is zero, the solution does not exhibit oscillations and approaches equilibrium as quickly as possible without overshooting.

Leveraging Computational Software for Differential Equations

Modern computational software offers an invaluable resource for solving complex differential equations that may not be solvable analytically. These tools use numerical methods to approximate solutions and provide visualizations that can aid in understanding the behavior of the system being studied.

For example, when working on an IVP like the one described in our exercise, computational software can generate a plot of the solution for different values of the damping constant \( \gamma \), allowing students to observe and analyse trends and relationships within the system's response over time.

System Behavior Analysis with Different Damping Constants

Analyzing the behavior of a system modeled by a differential equation is crucial in understanding the impact of different parameters on the system. In the context of the damped harmonic oscillator from our exercise, the damping constant \( \gamma \) plays a key role.

As \( \gamma \) increases, the system transitions from under-damped (oscillatory response) to critically damped (fastest non-oscillatory response) and eventually to over-damped (sluggish, non-oscillatory response). By plotting the solution for different values of \( \gamma \) - namely \( \gamma_{\text{crit}} \), \( 2\gamma_{\text{crit}} \), and \( 20\gamma_{\text{crit}} \) - one can observe these transitions and analyze how an increase in the damping constant affects the time it takes for the system to settle to a steady state.