Ph12a_s4_2009

# Ph12a_s4_2009 - Ph12a Solution Set 4 3.19 The absorptive...

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Unformatted text preview: Ph12a Solution Set 4 October 28, 2009 3.19 The absorptive and elastic amplitudes are given by A ab = F M Γ ω Γ 2 ω 2 + ( ω 2- ω 2 ) 2 (1) A el = F M ω 2- ω 2 Γ 2 ω 2 + ( ω 2- ω 2 ) 2 (2) The weak damping approximation is Γ / 2 << ω . This implies that ω ≈ ω , so we have ( ω 2- ω 2 ) 2 = ( ω + ω )( ω- ω ) = 2 ω ( ω- ω ) and the above expressions become: A ab = F M " Γ ω Γ 2 ω 2 (1 + 4( ω- ω ) 2 Γ 2 ) # (3) A el = F M " 2 ω ( ω- ω ) Γ 2 ω 2 (1 + 4( ω- ω ) 2 Γ 2 ) # (4) Defining x = ω- ω Γ / 2 , and once again using ω ≈ ω we therefore have: A ab = F M Γ ω 1 1 + x 2 (5) A el = F M Γ ω- x 1 + x 2 (6) Therefore in units of F M Γ ω , we have the desired forms of A ab and A el . 3.30 The equation of motion is Eq. (62) in the text: M d 2 ψ n dt 2 =- Mg l ψ n + K ( ψ n +1- ψ n )- K ( ψ n- ψ n- 1 ) . (7) We are to show that a solution of the form ψ ( z,t ) = A ( z )cos ωt with A ( z ) = A e- κz- e- κL e- κ ( L- z ) 1- e- 2 κL (8) 1 Ph12a Solutions Problem Set 4 October 28, 2009 satisfies the equations of motion, subject to the boundary conditions ψ (0 ,t ) = A cos ωt and ψ ( L,t ) = 0. We must check the solution satis- fies both the boundary conditions and the equation of motion. Boundary conditions —Substituting z = 0 and z = L into Eq. 8, we find that A (0) = A , A ( L ) = 0. Equation of motion —Substituting the solution into the RHS of the equa- tion of motion gives- Mg l ψ n + K ( ψ n +1- ψ n )- K ( ψ n- ψ n- 1 ) = cos ωt- Mg l A ( z ) + K [ A ( z + a ) + A ( z- a )- 2 A ( z...
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Ph12a_s4_2009 - Ph12a Solution Set 4 3.19 The absorptive...

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