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527solns13 - 642:527 SOLUTIONS ASSIGNMENT 13 FALL 2009...

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642:527 SOLUTIONS: ASSIGNMENT 13 FALL 2009 Section 19.2: 5. (a) The wire loop is treated as massless, so any force on it would cause infinite acceleration: there can be no (net) force on the loop. A nonzero u x ( L, t ) would lead to a nonzero vertical force, due to the string tension. Of course, the horizontal tension force is balanced by a force exerted on the loop by the wire. (b)–(d) The solution for (b) is given in the text; (c) and (d) are similar, using respectively half-range cosine and quarter-range cosine expansions: ( c ) : y ( x, t ) = V t ( d ) : y ( x, t ) = summationdisplay n =1 A n cos nπx 2 L sin nπct 2 L , A n = 4 nπc integraldisplay L 0 g ( x ) cos nπx 2 L dx. Note that there could be additional terms involving cos( nπct/L ) or cos( nπct/ 2 L ), but these vanish due to the condition that the initial displacement vanish. 6. Substituting y ( x, t ) = X ( x ) T ( t ) leads to X ′′ /X = ( T ′′ + aT ) / ( c 2 T ); as usual both sides must be constant, leading to the Sturm-Liouville problem X ′′ + λX = 0 , X (0) = X ( L ) = 0 , with known solutions λ n = ( nπ/L ) 2 , X n ( x ) = sin( nπx/L ) for n = 1 , 2 , . . . . Then T ′′ n + aT n + parenleftBig nπc L parenrightBig T N = 0 . This equation has characteristic polynomial r 2 + ar + ( nπc/L ) = 0 with (given the as- sumption 0 < a < 2 πc/L ) complex roots - μ n ± n , where μ n = a 2 , ν n = radicalBigg parenleftbigg 2 L parenrightbigg 2 - a 2 , and with solution T n ( t ) = e μ n t ( A n cos ν n t + B n sin ν n t ). Thus the soluiton is y ( x, t ) = summationdisplay n =1 sin( nπx L e μ n t ( A n cos ν n t + B n sin ν n t ) .
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