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A = 18 =1 C= 5 We have a string with length that is attached at both ends, both at x = 0 and x = 1. The differential equation and boundary conditions that describe the position u as a function of x and are given by Uu(x,t) = c²uzz(x,t), 0

A = 18 =1 C= 5 We have a string with length that is attached at both ends, both at x = 0 and x = 1. The differential equation and boundary conditions that describe the position u as a function of x and are given by Uu(x,t) = c²uzz(x,t), 0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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A =1B =1 C=5 We have a string with length that is attached at both ends, both at x = 0 and x = 1. The differential equation and boundary conditions that describe the position u as a function of x and are given by Uu(x,t) = c²u» (x,t), 0<x < l, 0<t<∞ 0<t<∞o, u(0,t) = u(l,t) = 0, 0<t< ∞, where the constant c is the wave speed of the string At time t = 0 we pull the middle of the string to a height h from the equilibrium position so that the position of the string is given by 0< x < u(x,0) = u;(x,0) = 0 All points on the string have a starting speed of zero at time t= 0, je for 0<x <1. (a) Write down the general solutions for position and speed of this string. Make simplifications as a result of the starting conditions (b) Calculate the coefficients of the general solutions so that you arrive at a solution for position and speed of the string as a function of I and h. To arrive at the solution, the following can be used (c) Explain, using the solution to the wave equation, why maximum speed is reached in the middle of the string (d) Find the wave velocity of the string given that the oscillation time T for an entire period is B seconds and the length I is your candidate number in centimeters (ABCcm). The value of B = 1. T= 5s and I= 115cm. (e) At what times does the string have the greatest speed?
Transcribed Image Text:A =1B =1 C=5 We have a string with length that is attached at both ends, both at x = 0 and x = 1. The differential equation and boundary conditions that describe the position u as a function of x and are given by Uu(x,t) = c²u» (x,t), 0<x < l, 0<t<∞ 0<t<∞o, u(0,t) = u(l,t) = 0, 0<t< ∞, where the constant c is the wave speed of the string At time t = 0 we pull the middle of the string to a height h from the equilibrium position so that the position of the string is given by 0< x < u(x,0) = u;(x,0) = 0 All points on the string have a starting speed of zero at time t= 0, je for 0<x <1. (a) Write down the general solutions for position and speed of this string. Make simplifications as a result of the starting conditions (b) Calculate the coefficients of the general solutions so that you arrive at a solution for position and speed of the string as a function of I and h. To arrive at the solution, the following can be used (c) Explain, using the solution to the wave equation, why maximum speed is reached in the middle of the string (d) Find the wave velocity of the string given that the oscillation time T for an entire period is B seconds and the length I is your candidate number in centimeters (ABCcm). The value of B = 1. T= 5s and I= 115cm. (e) At what times does the string have the greatest speed?
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