consider in the case of 1) internal resonance ω 2ω and 2) external resonance Ω ω and small damping
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1 consder n the cse o nternl resonnce nd externl resonnce Ω nd smll dmpng recll rom "Two_Degs_Frdm_.ppt" tht θ + μ θ + θ = θφ + cos Ω t + τ where = k α α nd φ + μ φ + φ = θ + cos Ω t where = α τ s constnt where θ represents the roll moton nd φ represents the ptch moton ntroduce bookkeepng prmeter ε nd rerrnge the EoMs to nd the smll dmpng ( μ = εμ = ε, : recognze the externl resonnce μ θ cos ( t θ + εμ θ + θ = θφ + ε cos Ω + τ ( t φ + ε φ + φ = + ε Ω (note: the coecents o the orcng terms were chosen so tht orcng, dmpng, nd nonlnerty wll nterct t the sme order
2 MMS : or smll-but-nte-mpltude motons, ssume pproxmtons n the ollowng orms ( t, ( T, T + ( T, T nd ( t, ( T, T + ( T, T θ ε εθ ε θ φ ε εφ ε φ then proceed s usul order ε: D θ + θ = 0 θ = A 0 ( T e T 0 + cc 0 ( DA A D φ + φ = 0 φ = e + 0 T0 A T cc ( DD D order ε : D θ + θ = θ + μ θ + θφ + cos Ω T + τ ( DD μ D θ cos( Ω D φ + φ = φ + φ + + T D θ + θ = + μ e D φ ( μ ( μ T 0 ( + ( ( Ω T + τ + AAe AAe + e + T T T0 T0 ΩT0 + = DA + A e + A e + AA + e + 0 or θ φ ( T 0 T0 : DA + A e wll produce seculr term; A A e wll produce smll-d becuse ; the remnng two terms wll produce nether seculr term not smll-dvsor term T0 T0 or φ : DA + μ A e wll produce seculr term; A e wll produce smll-dvsor term e cc cc vsor term ; ΩT0 becuse wll lso produce smll dvsor-term becuse Ω ; the remnng term wll not produce seculr term nor smll-dvsor term
3 to elmnte troublesome terms rom θ μ ( T0 T0, set D A + A e A A e = 0 to elmnte troublesome terms rom φ ( + μ T0 T0 ΩT0, set D A A e A e e = 0 to express smll-dvsor terms n the prmeters: Ω= + εσ nd = + εσ ( + ( ( T ( Ω scle T nsted o the scle T, ntroduce detunng T0 σ T e 0 e 0 0 T0 e 0 0 D A + μ A A A = D A + μ A A A = D A μ A A D e = A ( μ A σt + A e σt e 0 = to expedte the soluton, ntroduce polr coordntes: where nd β re rel β A = ( T e nd A = T β T T e ( β μ ( σ T β β + e = 4 ( + β + μ e 4 e = 0 ( σ T + β β ( σ T β let γ = σ T β nd γ = β β σ T
4 γ γ γ ( + β + μ e = 0 ( + β + μ e 4 4 e = 0 + sn = 0 + sn sn = 0 μ γ + μ γ γ cos = 0 + cos + cos = 0 β γ β γ γ 4 4 γ = σ β γ = β β σ 0 nd 0 γ σ γ γ γ γ γ γ σ = + cos + cos = cos cos + cos 4 4 one stedy-stte soluton when = 0 μ = sn γ, β = cos γ, γ = σ β = 0 μ ( μ + σ = = nd tn ( γ = σ μ + σ
5 ( T + β φ εφ = ε e + = ε e + T0 0 A cc cc φ ε μ σ ( + e ( + σ γ T T 0 + cc φ ε ε μ + ( Ω e ( Ω t γ + cc summry : ths rst pproxmton to the stedy-stte soluton s the soluton to the lner problem θ = 0 nd φ = cos Ω γ + ( Ω ε μ ( t
6 ( = = γ γ second stedy-stte soluton = = 0 when 0 nd 0 4μ = sn γ μ + sn γ = sn γ 4 μ + μ = sn γ γ = σ β = 0 σ = β nd γ = β β σ = 0 β = σ σ 3 β = cosγ 4 β + cosγ = cosγ = 0 4 ( σ σ γ σ γ γ 4 4 = cos + cos = cos ( ( σ σ σ σ = 4 = cosγ σ cosγ squre equtons nd, then dd the results = 4μ + σ σ
7 ( = = γ γ second stedy-stte soluton = = 0 when 0 nd 0 4μ = sn γ μ + sn γ = sn γ 4 μ + μ = sn γ γ = σ β = 0 σ = β nd γ = β β σ = 0 β = σ σ 3 β = cosγ 4 + cos = cos = 0 β γ γ 4 ( σ σ γ σ γ γ 4 4 = cos + cos = cos ( ( σ σ σ σ 4 = cos γ σ = cosγ squre eqns 3 nd 4, dd the results ( σ σ ( μ + μ + σ =
8 ( σ σ + + = ( μ μ σ s known; so ths s qudrtc equton n terms o the unknown summry: the two solutons re = 0 nd = μ σ ( + = Γ ± Γ nd = 4μ + σ σ where nd Γ = σ σ σ + μμ Γ = μσ μ σ σ note tht n, the mpltude o the only mode tht s drectly excted, does not depend on the mpltude o the exctton or on the dmpng o the drectly excted mode μ we determne the condtons under whch the cn exst
9 = Γ ± Γ = 4 + nd μ ( σ σ where nd Γ = σ σ σ + μμ Γ = μσ μ σ σ or soluton to exst, t s necessry, but not sucent, tht n ddton Γ Γ > 0, then one soluton exsts >Γ +Γ Γ < 0, then two solutons exst < Γ +Γ Γ < 0, then one soluton exsts > Γ +Γ
10 sturton phenomenon: comprson o experment nd theory derent chrcters or derent detunng prmeters, σ, σ roll st mode modl mpltudes modl mpltudes ptch ptch Mrshll s shp theoretcl hysteress supercrtcl nstblty subcrtcl nstblty nd mode Hddow s structure expermentl jumps orce orce
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