Chapter13-page3 - L 3 are excited. Energy of the String...

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Physics 235 Chapter 13 - 3 - The initial conditions are q x ,0 ( ) = 3 h L x , 0 ! x ! L 3 3 h 2 L L " x ( ) , L 3 ! x ! L # $ % % % % (1) ! q x ,0 ( ) = 0 (2) Because ! q x ,0 ( ) = 0 , all of the ! r vanish. The μ r are given by r = 6 h L 2 x sin r x L dx 0 L 3 " + 3 h L 2 L # x ( ) sin r x L dx L 3 L " = 9 h r 2 2 sin r 3 (3) We see that r = 0 for r = 3, 6, 9, etc. The displacement function is q x , t ( ) = 9 3 h 2 2 sin x L cos " 1 t + 1 4 sin 2 x L # cos 2 t $ 1 16 sin 4 x L cos 4 t $ % ( ) * (4) where r = r L # $ (5) The frequencies 3 , 6 , 9 , etc. are absent because the initial displacement at L 3 prevents that point from being a node. Thus, none of the harmonics with a node at
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Unformatted text preview: L 3 are excited. Energy of the String Since we know the displacement of the string, as function of x and t , q x , t ( ) = s e i s t sin s x L $ % & ( ) s * we can calculate the kinetic energy and the potential energy of the system....
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