By Khanh Chau Le

Half I. Linear thought: 1. unmarried oscillator -- 2. Coupled oscillators -- three. non-stop oscillators -- four. Linear waves -- half II. Nonlinear thought: five. self reliant unmarried oscillator -- 6. Non-autonomous unmarried oscillator -- 7. Coupled oscillators -- Nonlinear waves

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1) To use the energy method we write down the kinetic energy 1 1 K(˙x) = m1 x˙21 + m2 x˙22 , 2 2 and the potential energy 1 1 U(x) = k1 x21 + k2 (x2 − x1)2 . 1). 2 Coupled pendulums. Two pendulums are connected with each other by a spring of stiffness k (see Fig. 2). Derive the equations of motion for this system. 2. Because of the smallness of ϕ1 and ϕ2 , the magnitude of the spring force is equal to kl(ϕ2 − ϕ1 )/2, so the moment equation about A reads l2 m1 l 2 ϕ¨ 1 = ∑ Mz = −m1 gl ϕ1 + k (ϕ2 − ϕ1 ).

Thus, the Laplace transform of the convolution f ∗ g is equal to the product F(s)G(s) and vice versa. 31) yields τ x(τ ) = 0 g (t)xr (τ − t)dt. 28). Solution of initial-value problem. It turns out that the Laplace transform can also be used to find the solution of the initial-value problem x + 2δ x + x = 0, x(0) = x0 , x (0) = x0 . Indeed, applying the Laplace transform to this equation and observing that, due to the initial conditions, L [x ] = L [x ] = we obtain ∞ 0 ∞ 0 x e−sτ d τ = −x0 + sX(s), x e−sτ d τ = −x0 − sx0 + s2 X(s), (s2 + 2δ s + 1)X(s) = x0 + sx0 + 2δ x0.

5 2 (wj/wx)2 Fig. 8 Eigenfrequencies (ω1,2 /ωx )2 vs. ratio of uncoupled frequencies (ωϕ /ωx )2 at different coupling ratios (α /ωx )2 . The decay rates κ1,2 ω1,2 should be determined from the equation 4κω 4 − χω 3 − 2κ (ωx2 + ωϕ2 )ω 2 + χωωϕ2 = 0 giving κ 1 ω1 = χ (ω12 − ωϕ2 ) 4ω12 − 2(ωx2 + ωϕ2 ) , κ 2 ω2 = χ (ω22 − ωϕ2 ) 4ω22 − 2(ωx2 + ωϕ2 ) . Thus, the decay rates are positive and are of the same order as χ . 18) we may establish the relations between the amplitudes of vibrations. For s = (−κi ± i)ωi we have qˆ = xˆ ϕˆ =C (1 ± 2iκi)ωi2 − ωϕ2 , α22 i = 1, 2.