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M257-316Notes_Lecture25

# M257-316Notes_Lecture25 - Chapter 21 Lecture 25 Solution by...

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Chapter 21 Lecture 25 Solution by separation of variables Example 21.1 u tt = c 2 u xx 0 < x < L, t > 0 (21.1) BC: u (0 , t ) = 0 , u ( L, t ) = 0 (21.2) IC: u ( x, 0) = f ( x ) , u t ( x, 0) = g ( x ) (21.3) For a guitar string c = T 0 ρ 0 . Separate Variables u ( x, t ) = X ( x ) T ( t ) ¨ T ( t ) c 2 T ( t ) = X ( x ) X ( x ) = λ 2 (21.4) ¨ T ( t ) + λ 2 c 2 T ( t ) = 0 T ( t ) = c 1 cos( λct ) + c 2 sin( λct ) (21.5) X + λ 2 X = 0 X (0) = 0 = X ( L ) X ( x ) = A cos( λx ) + B sin λx X (0) = A = 0 X ( L ) = B sin λL = 0 λ n = L n = 1 , 2 , . . . X n = sin nπx L . 125

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Lecture 25 Solution by separation of variables Therefore u ( x, t ) = n =1 A n cos nπct L sin nπx L + B n sin nπct L sin nπx L (21.6) u ( x, 0) = n =1 A n sin nπx L = f ( x ) A n = 2 L L 0 f ( x ) sin ( nπx L ) (21.7) u t ( x, t ) = n =1 A n nπc L sin nπct L sin nπx L + B n nπc L cos nπct L sin nπx L (21.8) u t ( x, 0) = n =1 B n nπc L sin nπx L = g ( x ) B n ( nπc L ) = 2 L L 0 g ( x ) sin ( nπx L ) dx . (21.9) Therefore u ( x, t ) = n =1 A n cos nπct L + B n sin nπct L sin nπx L . (21.10) 21.1 Notes 1. Period and Frequency: cos nπc L ( t + T ) = cos nπct L provided nπcT L = 2 π (21.11) thus T n = 2 L c 1 n is the period (seconds per cycle) of mode n . f n = 1 T n = n c 2 L are the natural frequencies of vibration.
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