A string of length is stretched. We determine the axis coincides with the string so that the ends of the string are at the points x = 0 and x = L . The end of the string in x = 0 is unbound, while the end at x = L is bound. At first it is given a deviation, so that: y(x,t= 0) = d(1 - x/L) (d = a small value) Then the string is released, so that it vibrates and on the string propagates a wave with vines fast v . Using the orthogonality of the fundamental notes, find the wave function which propagates on the string y (x,t) Hint: since the end of the string at x = 0 is unbound, then: (in picture)
A string of length is stretched. We determine the axis coincides with the string so that the ends of the string are at the points x = 0 and x = L . The end of the string in x = 0 is unbound, while the end at x = L is bound. At first it is given a deviation, so that: y(x,t= 0) = d(1 - x/L) (d = a small value) Then the string is released, so that it vibrates and on the string propagates a wave with vines fast v . Using the orthogonality of the fundamental notes, find the wave function which propagates on the string y (x,t) Hint: since the end of the string at x = 0 is unbound, then: (in picture)
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 string of length is stretched. We determine the axis coincides with the string so that the ends of the string are at the points x = 0 and x = L . The end of the string in x = 0 is unbound, while the end at x = L is bound. At first it is given a deviation, so that:
y(x,t= 0) = d(1 - x/L) (d = a small value)
Then the string is released, so that it vibrates and on the string propagates a wave with vines fast v . Using the orthogonality of the fundamental notes, find the wave function which propagates on the string y (x,t)
Hint: since the end of the string at x = 0 is unbound, then: (in picture)
Transcribed Image Text:ду(х, t)
= 0
Əx
x=0
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