SolSec 6.2

# SolSec 6.2 - 6- 1 Chapter 6 Problems and Solutions Section...

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6- 1 Chapter 6 Problems and Solutions Section 6.2 (6.1 through 6.7) 6.1 Prove the orthogonality condition of equation (6.28). Solution: Calculate the integrals directly. For n = n , let u = n π x/l so that du = ( n π / l ) dx and the integral becomes l n udu l n uu l n nn l n n ππ π sin sin sin 2 0 0 1 2 1 4 2 1 2 1 4 40 2 =− −= where the first step used a table of integrals. For n m let u = π x/l so that d u = ( π / l ) dx and sin sin sin sin nx l mx l dx l mu nudu l l = 0 0 which upon consulting a table of integrals is lm n mn nm sin( ) () sin( ) + + = 22 0 .

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6- 2 6.2 Calculate the orthogonality of the modes in Example 6.2.3. Solution: One needs to show that XxXxd x m n Xt a x nm m n n () () , () s i n . =≠ = 0 0 1 for where σ But each mode X n ( x ) must satisfy equation (6.14), i.e. n n n X X 2 = (1) Likewise m m m X X 2 = (2) Multiply (1) by X m and integrate from 0 to l . Then multiply (2) by X n ( x ) and integrate from 0 to l . This yields ′′ =− XXdx XXd x x n l l mn m l l 2 0 0 2 0 0 Subtracting these two equations yields X X X X dx X x X x dx n m n m l l σσ 22 0 0 Integrate by parts on the left side to get + =−= X X dx X X dx X X X X Xl k Xl Xl k n l l m l l 0 0 00 0 from the boundary condition given by eq. (6.50). Thus n m l 0 0 =
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## This note was uploaded on 02/11/2010 for the course MECHANICAL ms316 taught by Professor Abduljaba during the Spring '09 term at Kalamazoo.

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SolSec 6.2 - 6- 1 Chapter 6 Problems and Solutions Section...

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