Question

Modify the problem in Example 2 in the text by replacing the given forcing function \(g(t)\) by $$ f(t)=\left[u_{s}(t)(t-5)-u_{5+k}(t)(t-5-k)\right] / k $$ (a) Sketch the graph of \(f(t)\) anple how it depends on \(k\). For what value of \(k\) is \(f(t)\) identical to \(g(t)\) in the example? (b) Solve the initial value problem $$ y^{\prime \prime}+4 y=f(t), \quad y(0)=0, \quad y^{\prime}(0)=0 $$ (c) The solution in part (b) depends on \(k,\) but for sufficiently large \(t\) the solution is always a simple harmonic oscillation about \(y=1 / 4\). Try to decide how the amplitude of this eventual oscillation depends on \(k .\) Then confirm your conclusion by plotting the solution for a few different values of \(k .\)

Short answer

The analysis and solution steps can be broken down as follows: 1. Sketch the graph: The graph of f(t) is zero for \(t<5\) and one for \(t \geq 5\). It does not depend on k directly, and the graph will not change as k is varied. 2. Determine the value of k where f(t) is identical to g(t): The only value of k where f(t) can be identical to g(t) is when k goes to infinity (\(k \rightarrow \infty\)). 3. Solve the initial value problem (IVP): The solution for the given IVP is: $$ y(t)=\frac{1}{4}[1-\cos(2t)]u_s(t)-\frac{1}{4}[1-\cos(2(t-(5+k)))] u_{5+k}(t) $$ 4. Determine the amplitude dependent on k: For large t values, the influence of k diminishes, and the oscillation amplitude will not be dependent on k. The amplitude of the oscillation is \(\frac{1}{4}\) and remains constant for different values of k.

Step-by-step solution

Sketching the graph of f(t)

To sketch the graph, we first need to understand the given expression for f(t): $$ f(t)=\frac{u_s(t)(t-5)-u_{5+k}(t)(t-5-k)}{k} $$ Now, consider two cases: 1. \(t<5\): In this case, \(u_s(t)=0\) and \(u_{5+k}(t)=0\). Hence, \(f(t)=0\). 2. \(t \geq 5\): In this case, \(u_s(t)=1\) and \(u_{5+k}(t)=1\). Thus, \(f(t)=\frac{(t-5)-(t-5-k)}{k}=\frac{k}{k}=1\) So, the graph of f(t) will be: - zero for \(t<5\) - one for \(t \geq 5\) To understand the dependence of f(t) on k, we can check how varying k affects the graph. However, the given f(t) expression does not depend on k directly, and the graph will not change as k is varied.

Determine the value of k where f(t) is identical to g(t)

For f(t) to be identical to g(t), the graph should match g(t)'s graph, and have the same properties as g(t). When looking at the graph of f(t), the only value of k where it can be identical to g(t) is when k goes to infinity (\(k \rightarrow \infty\)). This is because, in that case, the step function part of f(t) becomes similar to g(t). #b. Solving the initial value problem#

Solve the initial value problem (IVP)

To solve the given IVP: $$ y^{\prime \prime}+4 y=f(t), \quad y(0)=0, \quad y^{\prime}(0)=0 $$ We will first find the Laplace transform of the function: The Laplace transform of the given equation is: $$ Y(s)(s^2+4)=\frac{1-e^{-k s}}{s k} $$ Now, to obtain y(t), we need to find the inverse Laplace transform of Y(s). To simplify, we can perform partial fraction decomposition: $$ Y(s)(s^2+4) = \frac{A}{s} + \frac{Bs+C}{s^2+4} $$ The result is: $$Y(s)=\frac{1-e^{-k s}}{s k}\frac{s^2+4}{s^2+4}$$ Now, taking the inverse Laplace of Y(s), we get y(t): $$ y(t)=\mathcal{L}^{-1}\{Y(s)\}=\frac{1}{4}[1-\cos(2t)]u_s(t)-\frac{1}{4}[1-\cos(2(t-(5+k)))] u_{5+k}(t) $$ #c. Determine the amplitude of the eventual oscillation#

Determine the amplitude dependent on k

As mentioned in the problem statement, for sufficiently large t, the solution becomes a simple harmonic oscillation: $$ y(t) \approx \frac{1}{4} $$ This means that for large t values, the influence of k diminishes, and the oscillation amplitude will not be dependent on k. For large t, the first term in the solution dominates, making the amplitude of the oscillation independent of k. To confirm this, you can plot the solution found in Step 3 for various values of k and observe that the amplitude remains constant.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) is a differential equation that is solved by finding a function that satisfies both the equation and specific conditions given at the start, or 'initial' point. These conditions are usually expressed as function values and derivatives at a point, often at time zero. In the context of the exercise, we have:

• The differential equation: \( y'' + 4y = f(t) \)
• Initial conditions: \( y(0) = 0 \) and \( y'(0) = 0 \)

These conditions allow us to determine a unique solution for the function \( y(t) \) that evolves over time based on the forcing function \( f(t) \), which impacts the system after a certain delay. Studying IVPs helps us understand how specific conditions affect the behavior of dynamic systems as they change with time.
By solving the IVP, we apply methods such as the Laplace Transform to transform the equation into a more manageable form, solve for the unknown, and then revert back to the time domain, which gives us the function describing the system's behavior over time.

Laplace Transform

The Laplace Transform is a powerful mathematical tool used to simplify the process of solving differential equations, particularly those with initial conditions such as initial value problems. It is used to transform a time-domain function into the s-domain, which makes the equation algebraic and easier to handle. Here's how it works:

• Transforms a differential equation \( y'' + 4y = f(t) \) into an algebraic equation in terms of \( Y(s) \), the Laplace domain version of the function \( y(t) \).
• Initial conditions, such as \( y(0) = 0 \) and \( y'(0) = 0 \), are incorporated directly in the Laplace transformation process, simplifying the integration task throughout the formula.
• Utilizes properties like linearity and shifting that aid in breaking down complex signals.

In our exercise, we take the Laplace transform of both sides, resulting in:\[ Y(s)(s^2 + 4) = \frac{1 - e^{-ks}}{sk} \]After solving for \( Y(s) \), we apply the inverse Laplace Transform to retrieve \( y(t) \), the solution in the time domain.
The primary advantage of using the Laplace Transform is that it turns a problem involving calculus into one with simple algebra, making it easier to solve.

Simple Harmonic Oscillation

Simple Harmonic Oscillation is a type of periodic motion where the restoring force is directly proportional to the displacement. In differential equations, this is often represented by solutions that involve sine and cosine functions, describing oscillatory behavior over time.In the context of the given problem, after solving the initial value problem using the Laplace Transform, the resulting solution \( y(t) \) exhibits characteristics of simple harmonic oscillation as follows:

• For large values of \( t \), the solution simplifies to a simple harmonic oscillator around a mean position, which in this case is \( y = \frac{1}{4} \).
• This represents oscillations with a certain amplitude and frequency, independent of the parameter \( k \), when \( t \) becomes sufficiently large.

The realization of simple harmonic oscillation in this problem indicates that despite changes in initial conditions or parameters like \( k \), the system stabilizes to exhibit uniform oscillatory motion. This is crucial for studying systems like pendulums, springs, or any periodic motion, and helps predict long-term behavior without knowing every detail about the initial setup.
In real-world applications, understanding simple harmonic motion can guide us in designing systems that rely on predictable oscillatory behavior, like clocks or musical instruments.