SolSec3_3 - 3- 37Problems and Solutions Section 3.3...

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Unformatted text preview: 3- 37Problems and Solutions Section 3.3 (problems 3.26-3.32) 3.26Derive equations (3.24). (3.25) and (3.26) and hence verify the equations for the Fourier coefficient given by equations (3.21), (3.22) and (3.23). Solution: For n !m, integration yields: T!sinn"Ttsinm"Ttdt=sinn#m( )"Tt"T2n#m( )#sinn+m( )"Tt"T2n+m( )$%&&’())T=sinn#m( )2*T+,-./T$%&’()2n#m( )"T#sinn+m( )2*T+,-./T$%&’()2n+m( )"T=sinn#m( )2*( )$%’(2n#m( )"T#sinn+m( )2*( )$%’(2n+m( )"T=Since mand nare integers, the sine terms are 0, so this is equal to 0. Equation (3.24), for m= n: T!sin2n"Ttdt=12t#14n"Tsin 2n"Tt( )$%&’()T=T2#T8n*sin 2*2*T+,-./T$%&’()=T2#T8n*sin 4n*$%’(=T2Since nis an integer, the sine term is 0, so this is equal to T/2. So, T!sinn"Ttsinm"Ttdt=m#nT/ 2m=n$%&Equation (3.25), for m!n3- 38T!cosn"Ttcosm"Ttdt=sinn#m( )"Tt2n#m( )"T#sinn+m( )"Tt2$+m( )"T%&’’()**T=sinn#m( )2+T,-./1T%&’()*2n#m( )"T#sinn+m( )2+T,-./1T%&’()*2n+m( )"T=sinn#m( )2+( )%&()2n#m( )"T#sinn+m( )2+( )%&()2n+m( )"T=Since mand nare integers, the sine terms are 0, so this is equal to 0. Equation (3.25), for m= nbecomes: cos2T!n"Ttdt=12t+14n"Tsin 2n"Tt( )#$%&’(T=T2+T8n)sin 2n2)T*+,-./T#$%&’(=T2+T8n)sin 4n)#$&’=T2Since nis an integer, the sine term is 0, so this is equal to T/2. So, T!cosn"Ttcosm"Ttdt=m#nT/ 2m=n$%&Equation (3.26), form!n: T!cosn"Ttsinm"Ttdt=cosn#m( )"Tt2"Tn#m( )#cosn+m( )"Tt2"Tn+m( )$%&&’())T=cosn#m( )2*T+,-./T$%&’()2n#m( )"T#cosn+m( )2*T+,-./T$%&’()2n+m( )"T#12m#n( )"T+12m+n( )"T=cosn#m( )2*( )$%’(2n#m( )"T#cosn+m( )2*( )$%’(2n+m( )"T#12m#n( )"T+12m+n( )"T=Since nis an integer, the cosine term is 1, so this is equal to 0. 3- 39So, T!cosn"Ttsinm"Ttdt=Equation (3.26) for n = mbecomes: T!cosn"Ttsinn"Ttdt=12n"Tsin2n"Tt#$%&’(T=T4n)sin22)n=Thus T!cosn"Ttsinn"Ttdt=3- 403.27Calculate bnfrom Example 3.3.1 and show that bn= 0, n= 1,2,…,∞for the triangular force of Figure 3.12. Also verify the expression anby completing the integration indicated. (Hint: Change the variable of integration from tto x= 2πnt/T.) Solution: From Equation (3.23),bn=2TT!F t( )sinn"Ttdt. Computing the integral yields: bn=2TT/2!4Tt"1#$%&’(sinn)Ttdt+T/2T!1"4Tt"T2#$%&’(*+,-./sinn)Ttdt*+,,-.//bn=2T4TT/2!tsinn)Ttdt"T/2!sinn)Ttdt+3T/2T!...
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This note was uploaded on 03/31/2009 for the course MECHANICAL MAE351 taught by Professor J.g.lee during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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SolSec3_3 - 3- 37Problems and Solutions Section 3.3...

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