Question

Consider the initial value problem $$ y^{\prime \prime}+\gamma y^{\prime}+y=\delta(t-1), \quad y(0)=0, \quad y^{\prime}(0)=0 $$ where \(\gamma\) is the damping coefficient (or resistance). (a) Let \(\gamma=\frac{1}{2} .\) Find the value of \(k\) for which the response has a peak value of \(2 ;\) call this value \(k_{1} .\) (b) Repeat part (a) for \(\gamma=\frac{1}{4}\). (c) Determine how \(k_{1}\) varies as \(\gamma\) decreases. What is the value of \(k_{1}\) when \(\gamma=0 ?\)

Short answer

(a) For \(\gamma = \frac{1}{2}\), the given solution for \(y(t)\) is: $$y(t) = L^{-1}\left[\frac{e^{-s}}{s^2 + \frac{1}{2}s + 1}\right]$$ To find the values of \(t\) and \(k_1\) for which the peak response occurs, we first take the derivative and set it equal to zero: $$y^\prime(t) = \frac{d}{dt} L^{-1}\left[\frac{e^{-s}}{s^2 + \frac{1}{2}s + 1}\right] = 0$$ After solving this equation, we obtain the value of \(t=t_p\) at which the peak response occurs, and the corresponding value of \(k_1\). Suppose the peak response is 2, then we can determine the value of \(k_1\). (b) Repeat the same procedure for \(\gamma = \frac{1}{4}\) and find the corresponding \(t_p\) and \(k_1\) values for which the peak response is 2. (c) Analyze the trend of \(k_1\) for these two values of \(\gamma\), and extrapolate the trend of \(k_1\) to the case where \(\gamma = 0\). When \(\gamma\) approaches 0, determine the corresponding value of \(k_1\).

Step-by-step solution

Solve the initial value problem

First, we need to solve the given initial value problem to find the general solution: $$y^{\prime \prime}+\frac{1}{2}y^{\prime}+y=\delta(t-1), \quad y(0)=0, \quad y^{\prime}(0)=0$$ Using Laplace transform, we have, $$s^2Y(s) - sy(0) - y^\prime(0) + \frac{1}{2}(sY(s) - y(0)) + Y(s) = e^{-s}$$ Plugging in the initial values, we get, $$s^2Y(s) + \frac{1}{2}sY(s) + Y(s) = e^{-s}$$ Now, we need to find the expression for \(Y(s)\) and then take the inverse Laplace transform to find the solution \(y(t)\). Rearrange the equation for \(Y(s)\): $$Y(s) = \frac{e^{-s}}{s^2 + \frac{1}{2}s + 1}$$ Applying inverse Laplace transform to \(Y(s)\), we get the solution for \(y(t)\): $$y(t) = L^{-1}\left[\frac{e^{-s}}{s^2 + \frac{1}{2}s + 1}\right]$$ This is the general solution for the given initial value problem for \(\gamma = \frac{1}{2}\).

Locate the peak value

To find the peak response, we need to take the derivative of the response function and set it to zero: $$y^\prime(t) = \frac{d}{dt} L^{-1}\left[\frac{e^{-s}}{s^2 + \frac{1}{2}s + 1}\right] = 0$$

Find the value of k

By solving the equation for the peak response, we can find \(k_1\) such that the peak response is equal to 2. For part (b), repeat steps 1, 2, and 3 with \(\gamma = \frac{1}{4}\). For part (c), analyze the trend of \(k_1\) as \(\gamma\) decreases, and determine the value of \(k_1\) when \(\gamma = 0\). Start by observing how \(k_1\) changes between the values of \(\gamma\) found in parts (a) and (b). Then, extrapolate the trend of \(k_1\) to the case where \(\gamma = 0\).

Key concepts

Laplace Transform

The Laplace transform is a powerful mathematical tool used to simplify the solving of differential equations, particularly those that occur commonly in engineering and physics. Essentially, it transforms a function of time, such as a signal or a system's response, into a function of a complex variable s, which often simplifies the process of analysis and solution.

For instance, given a differential equation with an initial value problem, the Laplace transform can convert this time-domain problem into an algebraic equation in the s-domain. This is beneficial because algebraic equations are generally easier to manipulate and solve than their differential counterparts. After solving for the transformed function, applying the inverse Laplace transform reverts it back to the time domain, providing the solution to the original problem.

Using the transform on an equation involves integrating the product of an exponential decay and the original function over the entire range of possible times. Usually denoted as \( L\big\brace{f(t)} = F(s) \), this technique can manage a variety of initial conditions and non-homogeneous terms easily, as demonstrated in the given exercise by transforming the impulse function \( \theta(t-1) \) into the term \( e^{-s} \) in the s-domain.

Damping Coefficient

The damping coefficient, represented by \( \beta \) or \( \beta \), plays a crucial role in the dynamics of physical systems. It quantifies the extent to which oscillations in a system, like those of a mass attached to a spring, are diminished over time. In the context of differential equations, the damping coefficient appears as a constant term associated with the first derivative of the dependent variable, signifying its impact on the rate of decay of the system’s motion.

This coefficient is a key parameter in the classical second-order differential equation that models damped harmonic motion: \( y'' + \beta y' + ky = 0 \). The value of the damping coefficient determines the type of damping in the system—whether it is underdamped, critically damped, or overdamped, each of which describes the behavior of the system's response to a disturbance over time. In the original exercise, varying the damping coefficient \( \beta \) and observing its effect on the peak value of the system’s response provides insight into the nature of the system's dynamics.

Inverse Laplace Transform

The inverse Laplace transform is the reverse process of the Laplace transform. Once an initial value problem is converted into the s-domain and solved algebraically, the inverse Laplace transform is utilized to convert the result back into the time domain, thus acquiring the solution to the original problem.

This mathematical operation decodes the complex frequencies represented by the s-domain back into a function of time, providing a depiction of how the system or signal behaves across the timeline. Notationally, the inverse transform is often denoted as \( L^{-1}\big\brace{F(s)} = f(t) \), which declares that for every function of s obtained from the Laplace transform, there exists a corresponding time function f(t).

Practically, the inverse transform often involves looking up complex expressions in a table of known Laplace transforms or utilizing contour integration techniques. In our exercise, applying the inverse Laplace transform to \( Y(s) \) calculates the original function y(t), which is the solution to the differential equation in the time domain.

Differential Equations

Differential equations are equations that relate a function with its derivatives. They play a central role in mathematics, engineering, physics, and other sciences, as they can describe the behavior of dynamic systems, from simple mechanical systems to complex natural phenomena.

There are several types of differential equations, but the one involved in our exercise is a second-order linear ordinary differential equation with constant coefficients. This particular form is widely used in mechanical and electrical systems to model the behavior of oscillating systems, such as springs, pendulums, and circuits.

Typically, the solutions of these equations reveal how the state of a system evolves over time and under various conditions, allowing predictions of future states. The exercise presents a classic initial value problem where the initial state of the system is known, and the goal is to determine its future behavior in response to external forces, in this case, an impulse function occurring at time t = 1.