Problem 27 - Advanced Engineering Mathematics

Question

Consider the boundary-valueproblem introducod in theconstruction of the mathematical model for the shape of a rotating string: $$ T \frac{d^{2} y}{d x^{2}}+\rho w^{2} y=0, \quad y(0)=0, \quad y(L)=0 $$ For constant \(T\) and \(\rho\), define the mritical speeds of angular rotation \(\omega_{n}\) as the values of \(\omega\) for which the boundary-value pooblem has nontrivial solutions. Find the critical speeds \(\omega_{n}\) and the carrespanding deflections \(y_{n}(x)\).

Step-by-Step Solution

Answer

The critical speeds are \(\omega_n = \frac{n \pi \sqrt{T}}{L \sqrt{\rho}}\) and the deflections are \(y_n(x) = c_2 \sin\left(\frac{n \pi x}{L}\right)\).

Step 1: Identify the Differential Equation

Given the differential equation \(T \frac{d^{2} y}{d x^{2}} + \rho \omega^{2} y = 0\), identify it as a linear homogeneous second-order differential equation.

Step 2: General Solution of the Differential Equation

Assuming the solution \(y = e^{eta x}\), substitute into the differential equation to obtain the characteristic equation: \(T\beta^2 + \rho \omega^2 = 0\). Solving for \(\beta\) gives \(\beta = \pm i \frac{\omega \sqrt{\rho}}{\sqrt{T}}\). Thus, the general solution is \(y(x) = c_1 \cos\left(\frac{\omega \sqrt{\rho}}{\sqrt{T}} x\right) + c_2 \sin\left(\frac{\omega \sqrt{\rho}}{\sqrt{T}} x\right)\).

Step 3: Apply Boundary Conditions

Apply the boundary conditions \(y(0) = 0\) and \(y(L) = 0\) to the general solution. The boundary condition \(y(0) = 0\) requires \(c_1 = 0\), so \(y(x) = c_2 \sin\left(\frac{\omega \sqrt{\rho}}{\sqrt{T}} x\right)\). The boundary condition \(y(L) = 0\) implies \(c_2 \sin\left(\frac{\omega \sqrt{\rho}}{\sqrt{T}} L\right) = 0\), leading to \(\sin\left(\frac{\omega \sqrt{\rho}}{\sqrt{T}} L\right) = 0\).

Step 4: Determine Critical Speeds

The equation \(\sin\left(\frac{\omega \sqrt{\rho}}{\sqrt{T}} L\right) = 0\) is satisfied when \(\frac{\omega \sqrt{\rho}}{\sqrt{T}} L = n \pi\), where \(n\) is an integer. Solving for \(\omega\) gives the critical speeds: \(\omega_n = \frac{n \pi \sqrt{T}}{L \sqrt{\rho}}\).

Step 5: Find Corresponding Deflections

Using \(c_2 \sin\left(\frac{\omega \sqrt{\rho}}{\sqrt{T}} x\right)\) and substituting \(\omega_n\), the deflection is \(y_n(x) = c_2 \sin\left(\frac{n \pi x}{L}\right)\). The constant \(c_2\) can be chosen based on normalization or initial conditions.

Key Concepts

Differential Equations

Differential equations are equations that involve functions and their derivatives. Such equations are essential in modeling real-world phenomena, which include physics problems, engineering, and other applied sciences. They describe how a particular quantity changes with respect to another, usually over time or space.
In this specific exercise, the differential equation provided is \[ T \frac{d^{2} y}{d x^{2}} + \rho \omega^{2} y = 0 \]This is a second-order linear homogeneous differential equation. It is called 'second-order' because the highest derivative is second order, and 'homogeneous' because all the terms involve the unknown function or its derivatives. Here, the function \(y\) describes the shape of a rotating string, with \(T\) representing the tension and \(\rho\) the density of the string. Understanding and solving such equations is fundamental in determining dynamic behaviors and responses in systems.

Critical Speeds

Critical speeds in the context of this problem refer to specific angular velocities of the rotating string that cause it to assume certain stable shapes. These speeds are non-trivial solutions to the boundary value problem that ensure the system sustains oscillations without external forces.
To determine these speeds, we rely on the boundary conditions provided, such as \[ y(0) = 0 \quad \text{and} \quad y(L) = 0 \]Through these conditions and solving the associated characteristic equation, we recognize that the critical speeds can be expressed as \[ \omega_n = \frac{n \pi \sqrt{T}}{L \sqrt{\rho}} \]where \(n\) is a positive integer. These particular values of \(\omega\) correspond to the natural frequencies at which the string vibrates, making them pivotal in understanding the behavior of mechanical systems.

Deflection Solutions

The deflection solutions describe how the rotating string will flex or move at given critical speeds. By finding these solutions, we illustrate the shapes formed by the string when it is in equilibrium at these speeds. For our specific problem, the solution is presented once boundary conditions have been applied: \[ y_n(x) = c_2 \sin\left(\frac{n \pi x}{L}\right) \]This equation describes the deflection at the \(n\)-th critical speed \(\omega_n\). Here, \(c_2\) is a constant, which can be used for scaling based on certain initial or normalization conditions.

By analyzing deflection solutions, we can infer and predict how this mechanical system behaves under various operational conditions. Comprehending this is crucial in the design and control of systems subjected to rotational forces.

Linear Homogeneous Equations

Linear homogeneous equations are a category of differential equations that showcase the principle of superposition, meaning any linear combination of solutions is also a solution. These equations do not have a constant or non-zero term on the right-hand side. Our problem at hand is a linear homogeneous equation:\[ T \frac{d^{2} y}{d x^{2}} + \rho \omega^{2} y = 0 \]The lack of a constant term implies that if \(y(x)\) is a solution, then \(c \cdot y(x)\) is also a solution, for any constant \(c\). This characteristic is vital when applying boundary or initial conditions, like those for the rotating string.
In solving these equations, we typically find a general solution using characteristic equations derived from the assumption that the solution takes an exponential form. By substituting back into the original equation, we solve for parameters, allowing us to establish a solution applicable to various scenarios characterized by the system's specific boundary conditions.