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Unformatted text preview: 6-Chapter 6Problems and Solutions Section 6.2 (6.1 through 6.7)6.1Prove the orthogonality condition of equation (6.28).Solution:Calculate the integrals directly. For n= n, let u= nx/lso that du= (n/l)dxand the integral becomesln2=12-142=12-144- 0 =2where the first step used a table of integrals. For nmlet u= x/lso that du = (/l)dxand sinn = which upon consulting a table of integrals isl(-292(-29-(+292(+29= 0.16-6.2Calculate the orthogonality of the modes in Example 6.2.3.Solution:One needs to show thatXn(x)Xm(x)dx= 0 , (29=.1But each mode Xn(x) must satisfy equation (6.14), i.e.= -2(1)Likewise=2(2)Multiply (1) by Xmand integrate from 0 to l. Then multiply (2) by Xn(x) and integrate from 0 to l. This yields= -2= -2Subtracting these two equations yields-(29=2-2(29(29(29Integrate by parts on the left side to get-+-=(29(29-(29(29= 0from the boundary condition given by eq. (6.50). Thus2-2(29= 0....
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- Winter '09