Problem 33 - Advanced Engineering Mathematics

Question

Use the change of variables \(s=\frac{2}{\alpha} \sqrt{\frac{k}{m}} e^{-\alpha n / 2}\) to show that the differential equation of the aging spring \(m x^{\prime \prime}+k e^{-\alpha t} x=0\), \(\alpha>0\), becomes $$ s^{2} \frac{d^{2} x}{d s^{2}}+s \frac{d x}{d s}+s^{2} x=0 $$

Step-by-Step Solution

Answer

Transform the equation by substitution and simplify to derive the transformed differential equation.

Step 1: Identify the Original Differential Equation

The given differential equation is \( m x^{\prime \prime} + k e^{-\alpha t} x = 0 \). Our goal is to transform this using a change of variables.

Step 2: Substitute the Change of Variables

Replace \( t \) using the new variable \( s = \frac{2}{\alpha} \sqrt{\frac{k}{m}} e^{-\alpha t / 2} \). This implies \( e^{-\alpha t / 2} = \frac{\alpha s}{2} \sqrt{\frac{m}{k}} \) and thus \( e^{-\alpha t} = \left(\frac{\alpha s}{2} \sqrt{\frac{m}{k}}\right)^2 \). Substitute this in the original equation.

Step 3: Express Derivatives with Respect to the New Variable

Find \( \frac{dt}{ds} \) and \( \frac{d^2t}{ds^2} \) to replace derivatives of \( x \) with respect to \( t \) in terms of \( s \). Note, \( x^{\prime \prime} = \frac{d^2x}{dt^2} = \frac{d}{dt}(\frac{dx}{dt}) = \frac{d}{dt}(\frac{dx}{ds} \cdot \frac{ds}{dt}) \).

Step 4: Compute Derivatives for the Transformation

Calculate \( \frac{ds}{dt} = -\frac{\alpha s}{2} \) since \( s = \frac{2}{\alpha} \sqrt{\frac{k}{m}} e^{-\alpha t / 2} \). Then find \( \frac{dt}{ds} = -\frac{2}{\alpha s} \) and \( \frac{d}{dt}(\frac{dx}{dt}) = -(\frac{2}{\alpha s})^2\frac{d^2x}{ds^2} \).

Step 5: Substitute the Derived Expressions

Substitute the expressions back into the original differential equation: \( mx^{\prime\prime} = m (\frac{2}{\alpha s})^2 \frac{d^2x}{ds^2} \) and \( ke^{-\alpha t} x = k (\frac{\alpha s}{2} \sqrt{\frac{m}{k}})^2 x \).

Step 6: Simplify the Equation

Combine the substituted expressions to obtain \( s^2 \frac{d^2x}{ds^2} + s \frac{dx}{ds} + s^2 x = 0 \), which is the transformed differential equation.

Key Concepts

Differential Equations

Differential equations are a powerful tool in mathematics, often used to describe various phenomena in engineering, physics, and other sciences. At their core, a differential equation is an equation that relates a function with its derivatives. This typically involves rates of change and accumulations of quantities.

In our context, the original problem begins with the second-order differential equation:

• \( m x''(t) + k e^{-\alpha t} x(t) = 0 \).
This equation describes the motion of an aging spring, accounting for factors such as mass \( m \), spring constant \( k \), and a damping factor that exponentially decreases over time, represented by \( e^{-\alpha t} \).

Solving differential equations involves finding a function \( x(t) \) whose derivatives satisfy the equation. The complexity arises when additional elements, like time-dependent coefficients, are included, which makes solutions less straightforward and often necessitates specialized techniques or transformations.

Understanding differential equations also involves initial or boundary conditions, which provide additional context or constraints, leading to specific solutions that model real-world scenarios.

Transformation Techniques

Transformation techniques are mathematical strategies used to simplify complex differential equations, making them easier to solve. In this exercise, we employ a change of variables—a common transformation technique—by introducing a new variable \( s \) to transform the given equation into a more solvable form.

The specific transformation used here is:

• \( s = \frac{2}{\alpha} \sqrt{\frac{k}{m}} e^{-\alpha t / 2} \).

This substitution adjusts the variable \( t \) to \( s \), effectively simplifying the form of the differential equation. By expressing time \( t \) as a function of \( s \), we change the context of variation in the equation and allow the problematic exponential term \( e^{-\alpha t} \) to be controlled in terms of \( s \).

Transformation techniques are invaluable because they can reveal underlying simplicity in a system that might be obscured by its original formulation. They often turn nonlinear equations into linear ones, or at least into equations whose solutions are better understood. This technique highlights a key principle in mathematics: sometimes indirect paths (like a change of variables) offer more understanding and solvability than tackling the problem head-on.

Aging Spring Model

The aging spring model is a fascinating application of differential equations to describe a physical system. Here, we consider a spring-mass system where the spring's ability to function declines over time, represented mathematically by the decay factor \( e^{-\alpha t} \).

This model is particularly important in engineering and physics because it captures how materials or mechanisms might behave as they age or degrade. For instance, springs in a mechanical device might wear over time, resulting in reduced performance.

• Functionality: The model allows us to predict the position of the mass attached to the spring at any given time, taking the aging process into account.
• Parameters: The decay rate \( \alpha \) and the constants \( m \) and \( k \) provide insights into the specific characteristics of the spring involved.

In the aging spring model, the goal is to determine how the problem evolves over time and how variable changes (via techniques like transformation) can simplify tackling the analysis. Despite its complexity, the model encapsulates real-world mechanical behaviors, providing valuable insights into long-term performance and reliability of mechanical systems.