Chem Differential Eq HW Solutions Fall 2011 99

Chem Differential Eq HW Solutions Fall 2011 99 - 2 α j J 1...

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Section 6.3 The Hanging Chain 99 Solutions to Exercises 6.3 1. (a) The initial shape of the chain is given by the function f ( x )= - . 01( x - . 5) , 0 <x<. 5 , and the initial velocity of the chain is zero. So the solution is given by (10), with L = . 5 and B j = 0 for all j . Thus u ( x, t )= ± j =1 A j J 0 ² α j 2 x ³ cos ² ´ 2 g α j 2 t ³ . To compute A j , we use (11), and get A j = 2 J 2 1 ( α j ) µ . 5 0 ( - . 01)( x - . 5) J 0 ² α j 2 x ³ dx = - . 02 J 2 1 ( α j ) µ . 5 0 ( x - . 5) J 0 ² α j 2 x ³ dx Make the change of variables s = α j 2 x ,o r s 2 =2 α 2 j x ,so2 sds =2 α 2 j dx or dx = s α 2 j ds . Thus A j = - . 02 J 2 1 ( α j ) µ α j 0 ( . 5 α 2 j s 2 - . 5) J 0 ( s ) s α 2 j ds = - . 01 α 4 j J 2 1 ( α j ) µ α j 0 ( s 2 - α 2 j ) J 0 ( s ) sds = . 01 α 4 j J 2 1 ( α j ) 2 α 4 j α 2 j J 2 ( α j ) · = . 02 α 2 j J 2 1 ( α j ) J 2 ( α j ) , where we have used the integral formula (15), Section 4.3, with a = α = α j .W e can give our answer in terms of J 1 by using formula (6), Section 4.8, with p =1 , and x = α j . Since α j is a zero of
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Unformatted text preview: 2 α j J 1 ( α j ) = J ( α j ) + J 2 ( α j ) = J 2 ( α j ) . So A j = . 02 α 2 j J 2 1 ( α j ) 2 α j J 1 ( α j ) = . 04 α 3 j J 1 ( α j ) . Thus the solution is u ( x, t ) = ∞ ± j =1 . 04 α 3 j J 1 ( α j ) J ² α j √ 2 x ³ cos ² ´ 2 g α j 2 t ³ , where g ≈ 9 . 8 m/sec 2 . Going back to the questions, to answer (a), we have the normal modes u j ( x, t ) = . 04 α 3 j J 1 ( α j ) J ² α j √ 2 x ³ cos ² ´ 2 g α j 2 t ³ . The frequency of the j th normal mode is ν j = ´ 2 g α j 4 π ....
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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