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ODWE_11-23

ODWE_11-23 - Tutorial ODWE One-Dimensional W Equation ave...

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Tutorial ODWE: One-Dimensional Wave Equation Thus, @ 2 f @x 2 ¡ 1 c 2 @ 2 f @t 2 = d 2 f du 2 ¡ d 2 f du 2 = 0 : The proof is similar for f ( u + ) : Exercise 10 . See, for example, Fishbane, Gasiorowicz and Thornton, ‘‘Physics for Scientists and Engineers (Extended)’’ , Second Edition, Prentice Hall, 1996, pp. 380 - 381. Exercise 11 . Taking partial derivatives, we have @ 2 y ( x; t ) @x 2 = T ( t ) d 2 X dx 2 ; @ 2 y ( x; t ) @t 2 = X ( x ) d 2 T dt 2 : Substituting these expressions into Eqn. 6, we obtain T ( t ) d 2 X dx 2 ¡ 1 À 2 X ( x ) d 2 T dt 2 = 0 : We now divide both sides by y ( x; t ) = X ( x ) T ( t ) to get 1 X d 2 X dx 2 ¡ 1 À 2 1 T d 2 T dt 2 = 0 : Exercise 12 . We simply multiply both sides by X : d 2 X dx 2 = ¡ k 2 X; which has the solution sets X ( x ) = ½ cos kx sin kx ¾ or ½ exp( ikx ) exp( ¡ ikx ) ¾ : Exercise 13 . Since the boundary conditions are periodic in nature (a zero occurs for two different values of x ) we will take the general solution to be X ( x ) = A cos kx + B sin kx: The condition X (0) = 0 yields 0 = A ¢ 1 + B ¢ 0 = ) A = 0 ; leaving us with X ( x ) = B sin kx: The second condition, X ( L ) = 0 then requires either B = 0

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ODWE_11-23 - Tutorial ODWE One-Dimensional W Equation ave...

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