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A guitar string is clamped at both ends. For the purpose of this problem we may consider it to have a length, L = x, and a wave speed, c = 1. Show that, u(x, t) = sin(x) cos(t) – } sin(3r) cos(3t) + sin(5x) cos(5t) (a) Satisfies both boundary conditions (clamped at both ends) (b) Is a solution to the wave equation (c) (Intermediate) Satisfies the initial condition if the guitar string is plucked - that is that (x,0) = 0

A guitar string is clamped at both ends. For the purpose of this problem we may consider it to have a length, L = x, and a wave speed, c = 1. Show that, u(x, t) = sin(x) cos(t) – } sin(3r) cos(3t) + sin(5x) cos(5t) (a) Satisfies both boundary conditions (clamped at both ends) (b) Is a solution to the wave equation (c) (Intermediate) Satisfies the initial condition if the guitar string is plucked - that is that (x,0) = 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 guitar string is clamped at both ends. For the purpose of this problem we may consider it to have a length, L = x, and a wave speed, c = 1. Show that, u(x, t) = sin(x) cos(t) – sin(3r) cos(3t) + sin(5x) cos(5t) (a) Satisfies both boundary conditions (clamped at both ends) (b) Is a solution to the wave equation (c) (Intermediate) Satisfies the initial condition if the guitar string is plucked - that is that (x, 0) = 0 The "period", T, of the solution is the shortest time it takes for the guitar string to return to its original shape, i.e., u(x, KT) = ...= u(x, 27') = u(x, T') = u(x, 0) (d) The period for our string is, T = 27. Show this by demonstrating that, u(r, T) = u(x, 0) (e) If we change the wave speed to c = 2, our solution becomes, u(x, t) = sin(x) cos(2t) – sin(3r) cos(6t) + sin(5x) cos(10t) What is the period now? – note, you may want to check the argument used to apply clamped conditions u(0, t) = u(x,t) = 0
Transcribed Image Text:A guitar string is clamped at both ends. For the purpose of this problem we may consider it to have a length, L = x, and a wave speed, c = 1. Show that, u(x, t) = sin(x) cos(t) – sin(3r) cos(3t) + sin(5x) cos(5t) (a) Satisfies both boundary conditions (clamped at both ends) (b) Is a solution to the wave equation (c) (Intermediate) Satisfies the initial condition if the guitar string is plucked - that is that (x, 0) = 0 The "period", T, of the solution is the shortest time it takes for the guitar string to return to its original shape, i.e., u(x, KT) = ...= u(x, 27') = u(x, T') = u(x, 0) (d) The period for our string is, T = 27. Show this by demonstrating that, u(r, T) = u(x, 0) (e) If we change the wave speed to c = 2, our solution becomes, u(x, t) = sin(x) cos(2t) – sin(3r) cos(6t) + sin(5x) cos(10t) What is the period now? – note, you may want to check the argument used to apply clamped conditions u(0, t) = u(x,t) = 0
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