Math Problem Statement
Find the displacement of a string stretched between two fixed points at a distance 2 π apart when the string is initially at rest in equilibrium position and points of the string are given initial velocities π£ where π£ = { π₯π , ππ 0 < π₯ < π 2 π β π₯ π , ππ π < π₯ < 2 π
Solution
To solve for the displacement of a string stretched between two fixed points, we need to consider the initial conditions given by the problem. The string is initially at rest in an equilibrium position and is given an initial velocity described by the piecewise function:
\begin{cases} \frac{x}{a}, & \text{if } 0 < x < a, \\ \frac{2a - x}{a}, & \text{if } a < x < 2a. \end{cases}$$ ### Step 1: Understand the Problem The problem involves a wave equation for a string fixed at both ends. The general solution to the wave equation can be expressed as: $$u(x, t) = f(x - ct) + g(x + ct),$$ where $$c$$ is the wave speed, $$f$$ and $$g$$ are arbitrary functions determined by the initial conditions. The boundary conditions are $$u(0, t) = 0$$ and $$u(2a, t) = 0$$, since the string is fixed at $$x = 0$$ and $$x = 2a$$. ### Step 2: Initial Conditions 1. **Initial Displacement**: The string is initially at rest in the equilibrium position, so $$u(x, 0) = 0.$$ 2. **Initial Velocity**: The initial velocity $$\frac{\partial u}{\partial t}(x, 0)$$ is given by the function $$v(x)$$: $$\frac{\partial u}{\partial t}(x, 0) = v(x).$$ ### Step 3: Solve the Wave Equation Using D'Alembert's solution for the wave equation: $$u(x, t) = \frac{1}{2}\left[ f(x - ct) + f(x + ct) \right] + \frac{1}{2c}\int_{x - ct}^{x + ct} v(s) \, ds.$$ Given that the initial displacement is zero, the solution simplifies to: $$u(x, t) = \frac{1}{2c}\int_{x - ct}^{x + ct} v(s) \, ds.$$ ### Step 4: Compute the Integral for $$u(x, t)$$ The integral for $$u(x, t)$$ can be computed by splitting the integral based on the piecewise definition of $$v(x)$$: $$u(x, t) = \frac{1}{2c} \left[ \int_{x - ct}^{a} \frac{s}{a} \, ds + \int_{a}^{x + ct} \frac{2a - s}{a} \, ds \right].$$ Let's evaluate these integrals. 1. **For $$0 < x < a$$:** $$u(x, t) = \frac{1}{2c} \left[ \int_{x-ct}^{x+ct} \frac{s}{a} \, ds \right].$$ 2. **For $$a < x < 2a$$:** $$u(x, t) = \frac{1}{2c} \left[ \int_{x-ct}^{a} \frac{s}{a} \, ds + \int_{a}^{x+ct} \frac{2a - s}{a} \, ds \right].$$ ### Step 5: Perform Integration Now we evaluate the integrals for both cases: 1. **For $$0 < x < a$$:** \[ u(x, t) = \frac{1}{2c} \left[ \frac{1}{a} \left( \frac{s^2}{2} \right) \Bigg|_{x-ct}^{x+ct} \right] = \frac{1}{2ac} \left[ \frac{(x + ct)^2 - (x - ct)^2}{2} \right]. \] Simplifying this expression will give the displacement. 2. **For $$a < x < 2a$$:** For this interval, you would need to evaluate the two separate integrals considering the bounds, and then sum the results to obtain the displacement $$u(x, t)$$. ### Final Displacement Solution: The final displacement expression will be piecewise, similar to the initial velocity. The specific forms of the integrals, once evaluated and simplified, give the complete expression for $$u(x, t)$$. Would you like a more detailed integration step or any specific part of the solution to be expanded upon? Here are some relative questions you might explore: 1. How does the boundary condition affect the solution of the wave equation? 2. What is the physical interpretation of the wave speed $$c$$ in the context of the string? 3. Can this method be applied to strings with different initial conditions, such as different initial displacements? 4. How would the solution change if the string were not fixed at both ends? 5. How does the displacement vary with time for different points on the string? **Tip:** Understanding the integral limits and how they interact with the piecewise function is crucial in solving wave equations with non-uniform initial conditions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Wave Equations
Partial Differential Equations
Initial and Boundary Value Problems
Piecewise Functions
Formulas
Wave equation: u(x, t) = f(x - ct) + g(x + ct)
D'Alembert's solution: u(x, t) = (1/2) * [f(x - ct) + f(x + ct)] + (1/2c) * β«_{x - ct}^{x + ct} v(s) ds
Theorems
D'Alembert's Solution of the Wave Equation
Suitable Grade Level
Undergraduate level - Advanced Calculus or Mathematical Physics
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