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Unformatted text preview: 1 & Nonresonant Hard Excitation We still need to use K as O i.e. Let the solution be Substituting into the equation gives The solution for O( ) is 2 1 1 1 u(t; ) u (T ,T ) u (T ,T ) O( ) = + + ( ) 2 3 u u (2 u u ) Kcos t + + + = && & 2 2 1 2 2 3 1 1 1 O( ) : D u u K cos T O( ) : D u u 2D D u 2 D u u + = + =   1 O( ) : u A(T )exp(i T ) exp(i T ) C.C. = + + We go to the next order equation: Clearly, there are many ways in which secular terms or small denominators may arise in the solution for u 1 (T , T 1 ). 1 2 2 3 1 1 1 2 2 3 3 2 2 2 2 2 O( ) : D u u 2D D u 2 D u u [2i (A A) 6 A 3 A A]exp(i T ) {A exp(3i T ) exp(3i T ) 3A exp[i(2 )T ] 3A exp[i( 2 )T ] 3A exp[i( 2 )T ] 3A exp[i( 2 )T ]} [2i 3 6 + =   =  + + +  + + + +  + + + +   + + AA]exp(i T ) C.C. + 2 & Whatever may be the situation in terms of the frequency of excitation, secular terms always exist due to the first term on the right hand side. Thus, in the nonresonant case, we have Let We get the amplitude and phase equations: The first approximation to solution is: 2 2 1 dA S.T : 2i A 6 A 3 A A dT & + + + = 1 A a exp(i ) 2 = 2 2 1 a a, a 3 ( a )a 8 =  = + 2 2 K u a cos( t ) cos t O( ) ( ) = + + +  Superharmonic Resonance K is again large i.e., this is also a case of (non resonant) or hard excitation! Solution : (does not blow up!) We now go to the next order: i T 1 Step 2 : u A(T )e C.C exp(i T ) C.C = + + + 2 1 1 1 Step1: u(t; ) u (T ,T ) u (T ,T ) O( ) = + + O( ) 2 2 1 1 where K( ) 2 =  3 & The next order equation is: (Note: same equation as in nonresonant case) Clearly, there are many ways in which secular terms or small denominators may arise in the solution for u 1 (T , T 1 ). 1 2 2 3 1 1 1 2 2 3 3 2 2 2 2 2 O( ) : D u u 2D D u 2 D u u [2i (A A) 6 A 3 A A]exp(i T ) {A exp(3i T ) exp(3i T ) 3A exp[i(2 )T ] 3A exp[i( 2 )T ] 3A exp[i( 2 )T ] 3A exp[i( 2 )T ]} [2i 3 6 + =   =  + + +  + + + +  + + + +   + + AA]exp(i T ) C.C. + Step 3: Generate secular terms in equation Then, and the terms generating secular terms have coefficient Let 1 O( ) (the problem arises when 3 , 3 ) 3 3 = + & = + 1 3 T ( )T T T T T = + = + = + 2 2 1 3 1 dA S.T : 2i A 6 A 3 A A dT exp(i T ) additional term due to sup erharmonic resonance + + + + = & 1 A a exp(i ) 2 = 4 & Then, the equations for a and are: If one defines, We get autonomous equations: Step 4: Equilibria: 2 2 1 Recall that u a cos(3 t ) K( ) cos t O( ) =  +  + 3 2 2 3 1 1 1 a a sin( T ), a...
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This note was uploaded on 12/28/2011 for the course ME 580 taught by Professor Na during the Fall '10 term at Purdue UniversityWest Lafayette.
 Fall '10
 NA

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