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Consider an elastic string of length \(L .\) The end \(x=0\) is held fixed while the end \(x=L\) is free; thus the boundary conditions are \(u(0, t)=0\) and \(u_{x}(L, t)=0 .\) The string is set in motion with no initial velocity from the initial position \(u(x, 0)=f(x),\) where $$ f(x)=\left\\{\begin{array}{ll}{1,} & {L / 2-12)} \\ {0,} & {\text { otherwise. }}\end{array}\right. $$ (a) Find the displacement \(u(x, t) .\) (b) With \(L=10\) and \(a=1\) plot \(u\) versus \(x\) for \(0 \leq x \leq 10\) and for several values of \(t .\) Pay particular attention to values of \(t\) between 3 and \(7 .\) Observe how the initial disturbance is reflected at each end of the string. (c) With \(L=10\) and \(a=1\) plot \(u\) versus \(t\) for several values of \(x .\) (d) Construct an animation of the solution in time for at least one period. (e) Describe the motion of the string in a few sentences.

Short Answer

Expert verified
Based on the given elastic string length L, boundary conditions, and initial position f(x), we found the displacement function u(x,t) to be: $$ u(x,t)=\sum_{n=1}^{\infty}\frac{2}{n\pi}\left[\cos\left(\frac{n\pi(L/2-1)}{L}\right)-\cos\left(\frac{n\pi(L/2+1)}{L}\right)\right] \sin\left(\frac{n\pi x}{L}\right) \cos\left(an\frac{\pi t}{L}\right) $$ By plotting the displacement u(x,t) with respect to x and t, we discovered that the initial disturbance of the string travels from the middle towards the ends, reflecting back and forth between the fixed ends. The wave motion remains unchanged over time as the string has no damping mechanism.

Step by step solution

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01

Find Fourier series of function f(x)

To find the function u(x,t), we first need to find the Fourier sine series of the initial position f(x). The Fourier sine series representation is given by: $$ f(x) = \sum_{n=1}^{\infty} B_n \sin \left(\frac{n \pi x}{L} \right) $$ Where: $$ B_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx $$ Now, we will calculate B_n for f(x).
02

Calculate B_n for f(x)

As f(x) is piecewise-defined, we will split the integral for B_n into two parts: $$ B_n = \frac{2}{L} \int_{L/2-1}^{L/2+1} 1 \cdot \sin\left(\frac{n\pi x}{L}\right) dx = \frac{2}{L} \bigg[ \left. -\frac{L}{n\pi}\cos\left(\frac{n\pi x}{L}\right)\right|_{L/2-1}^{L/2+1} \bigg] $$ Calculate B_n by solving the integral: $$ B_n = \frac{2}{n\pi}\left[\cos\left(\frac{n\pi(L/2-1)}{L}\right)-\cos\left(\frac{n\pi(L/2+1)}{L}\right)\right] $$ Now, we will find u(x,t) using B_n.
03

Solve the wave equation

To find u(x,t), we will use the following formula of the wave equation for a fixed-end string: $$ u(x,t) = \sum_{n=1}^{\infty} B_n \sin \left(\frac{n \pi x}{L} \right)\cos\left(an\frac{\pi t}{L}\right) $$ Replacing f(x) with the B_n value calculated in Step 2: $$ u(x,t)=\sum_{n=1}^{\infty}\frac{2}{n\pi}\left[\cos\left(\frac{n\pi(L/2-1)}{L}\right)-\cos\left(\frac{n\pi(L/2+1)}{L}\right)\right] \sin\left(\frac{n\pi x}{L}\right) \cos\left(an\frac{\pi t}{L}\right) $$ Now, we can use the above formula to plot u(x,t) for different values of t and x with L = 10 and a = 1. #b - Plot u versus x for 0 <= x <= 10 and several values of t# To plot u versus x for 0 <= x <= 10, we will consider values of t between 3 and 7, with L = 10 and a = 1. Use the formula of u(x,t) derived in the previous step and a graphing tool to visualize the wave pattern. #c - Plot u versus t for several values of x# To plot u(x,t) with L = 10 and a = 1 for several values of x, use the obtained formula for u(x,t) and a graphing tool to visualize the wave with respect to time. #d - Construct an animation of the solution in time for at least one period# Using the u(x,t) formula and animation tools, create an animation of the wave for at least one period of the wave. This will help visualize the behavior of the wave over time. #e - Describe the motion of the string# After observing the plots and animations, it can be concluded that the initial disturbance of the string reflects back and forth between the ends of the string with fixed end reflections. The initial bump travels from the middle towards the ends, with no energy loss over time, as the string has no damping mechanism.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Equation Solutions
The wave equation is a fundamental partial differential equation that describes waves' behavior, such as vibrations on a string. This two-variable equation involves time \( t \) and spatial position \( x \). A simple form of the wave equation for a one-dimensional string is: \[\frac{\partial^2 u}{\partial t^2} = a^2 \frac{\partial^2 u}{\partial x^2}\] where \( u(x, t) \) represents the displacement of the wave, and \( a \) is the wave speed. To solve a wave equation, we rely on methods like separation of variables, which involves breaking down the problem into simpler parts that can be solved independently and then recombined. This technique often leads to solutions in the form of a series, such as the Fourier series, representing the wave as a sum of sine and cosine functions.
Boundary Conditions
Boundary conditions are rules applied at the endpoints of the string that help determine its behavior. They are critical to finding a unique solution to the wave equation. For an elastic string of length \( L \), boundary conditions might be given as:
  • Fixed end at \( x = 0 \), meaning \( u(0, t) = 0 \)
  • Free end at \( x = L \), meaning \( \frac{\partial u}{\partial x}(L, t) = 0 \)
These boundary conditions specify what happens at the edges of the string as it oscillates. For example, a fixed endpoint cannot move, while a free end has zero slope, akin to the string having no directional constraint at that point. By applying these conditions, we specify the wave behaviors precisely at the string's ends.
Initial Conditions
Initial conditions describe the state of the system at the start of observation, i.e., at \( t = 0 \). They help further clarify the specific wave scenario of interest. A typical initial condition for a wave equation is the string's shape and velocity at the initial time:
  • Initial position: \( u(x, 0) = f(x) \)
  • Initial velocity: Often set to zero, such as \( \frac{\partial u}{\partial t}(x, 0) = 0 \)
In the given example, the string starts at a specific shape with no initial velocity. The function \( f(x) \) describes the initial bump, with values of 1 between \( L/2-1 \) and \( L/2+1 \) and zero elsewhere. These initial conditions will determine how the wave propagates and evolves over time in the solution.
Fourier Sine Series
A Fourier sine series is a way to represent a function as a sum of sinusoidal functions, specifically sine waves. This representation is powerful for solving problems with specific boundary conditions, particularly when the endpoints are set in a certain way, like a fixed end.The Fourier sine series of a function \( f(x) \) over an interval \( [0, L] \) is expressed as: \[f(x) = \sum_{n=1}^{\infty} B_n \sin \left(\frac{n \pi x}{L} \right)\]The coefficients \( B_n \) are determined by the formula: \[B_n = \frac{2}{L} \int_{0}^{L} f(x) \sin \left(\frac{n \pi x}{L} \right) dx\]In the context of solving the wave equation, the Fourier sine series is used to decompose the initial displacement \( f(x) \) into harmonics—these form the basis of the wave solution over time. By calculating \( B_n \) for given \( f(x) \), we obtain a series that captures all necessary frequencies to describe the string's motion accurately, respecting both the boundary and initial conditions.

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Most popular questions from this chapter

Carry out the following steps. Let \(L=10\) and \(a=1\) in parts (b) through (d). (a) Find the displacement \(u(x, t)\) for the given initial position \(f(x) .\) (b) Plot \(u(x, t)\) versus \(x\) for \(0 \leq x \leq 10\) and for several values of \(t\) between \(t=0\) and \(t=20\). (c) Plot \(u(x, t)\) versus \(t\) for \(0 \leq t \leq 20\) and for several values of \(x .\) (d) Construct an animation of the solution in time for at least one period. (e) Describe the motion of the string in a few sentences. \(f(x)=8 x(L-x)^{2} / L^{3}\)

In each of Problems 19 through 24 : (a) Sketch the graph of the given function for three periods. (b) Find the Fourier series for the given function. (c) Plot \(s_{m}(x)\) versus \(x\) for \(m=5,10\), and 20 . (d) Describe how the Fourier series seems to be converging. $$ f(x)=\left\\{\begin{array}{lr}{x+2,} & {-2 \leq x < 0,} \\ {2-2 x,} & {0 \leq x < 2}\end{array} \quad f(x+4)=f(x)\right. $$

Consider the problem $$ \begin{aligned} \alpha^{2} u_{x x}=u_{t}, & 00 \\ u(0, t)=0, \quad u_{x}(L, t)+\gamma u(L, t)=0, & t>0 \\ u(x, 0)=f(x), & 0 \leq x \leq L \end{aligned} $$ (a) Let \(u(x, t)=X(x) T(t)\) and show that $$ X^{\prime \prime}+\lambda X=0, \quad X(0)=0, \quad X^{\prime}(L)+\gamma X(L)=0 $$ and $$ T^{\prime}+\lambda \alpha^{2} T=0 $$ where \(\lambda\) is the separation constant. (b) Assume that \(\lambda\) is real, and show that problem (ii) has no nontrivial solutions if \(\lambda \leq 0\). (c) If \(\lambda>0\), let \(\lambda=\mu^{2}\) with \(\mu>0 .\) Show that problem (ii) has nontrivial solutions only if \(\mu\) is a solution of the equation $$ \mu \cos \mu L+\gamma \sin \mu L=0 $$ (d) Rewrite Eq. (iii) as \(\tan \mu L=-\mu / \gamma .\) Then, by drawing the graphs of \(y=\tan \mu L\) and \(y=-\mu L / \gamma L\) for \(\mu>0\) on the same set of axes, show that Eq. (iii) is satisfied by infinitely many positive values of \(\mu ;\) denote these by \(\mu_{1}, \mu_{2}, \ldots, \mu_{n}, \ldots,\) ordered in increasing size. (e) Determine the set of fundamental solutions \(u_{n}(x, t)\) corresponding to the values \(\mu_{n}\) found in part (d).

This problem indicates a proof of convergence of a Fourier series under conditions more restrictive than those in Theorem \(10.3 .1 .\) (a) If \(f\) and \(f^{\prime}\) are piecewise continuous on \(-L \leq x

In each of Problems 19 through 24 : (a) Sketch the graph of the given function for three periods. (b) Find the Fourier series for the given function. (c) Plot \(s_{m}(x)\) versus \(x\) for \(m=5,10\), and 20 . (d) Describe how the Fourier series seems to be converging. $$ f(x)=x^{2} / 2, \quad-2 \leq x \leq 2 ; \quad f(x+4)=f(x) $$

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