11. An ideal elastic string of length L is fixed at its right end, but its left end is attached to a ring that is free to slide up and down a vertical pole. One can show that under these assumptions the tangent line to the free end will always be horizontal. Separate variables and describe the motion of the string at any time t> 0, given any initial shape and (vertical) velocity. Positioning the string along the interval [0, L] of the x-axis, solve the boundary value problem by separation of variables. Utt = c²uxx, 0 < x 0 ur(0,t) = u(L,t)=0, t>0 u(x, 0) = f(x), ut(x,0) = g(x), 0 < x < L

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 91E
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11. An ideal elastic string of length L is fixed at its right end, but its left end is
attached to a ring that is free to slide up and down a vertical pole. One can
show that under these assumptions the tangent line to the free end will always
be horizontal. Separate variables and describe the motion of the string at any
time t > 0, given any initial shape and (vertical) velocity. Positioning the
string along the interval [0, L] of the x-axis, solve the boundary value problem
by separation of variables.
Utt = c²uxx, 0 < x <L, t>0
ur(0,t) = u(L,t)=0, t>0
u(x, 0) = f(x), ut(x,0) = g(x), 0 < x < L
Transcribed Image Text:11. An ideal elastic string of length L is fixed at its right end, but its left end is attached to a ring that is free to slide up and down a vertical pole. One can show that under these assumptions the tangent line to the free end will always be horizontal. Separate variables and describe the motion of the string at any time t > 0, given any initial shape and (vertical) velocity. Positioning the string along the interval [0, L] of the x-axis, solve the boundary value problem by separation of variables. Utt = c²uxx, 0 < x <L, t>0 ur(0,t) = u(L,t)=0, t>0 u(x, 0) = f(x), ut(x,0) = g(x), 0 < x < L
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