Hence t 2 2 c ε a 1 a 2 x 2 3 4 ε a 2 x 2 d x 2 2 c

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Hence T 2 2 c ε a 0 1 (a 2 x 2 ) + 3 4 ε (a 2 x 2 ) d x = 2 2 c ε sin 1 (x/a) + 3 8 ε { x (a 2 x 2 ) + a 2 sin 1 (x/a) } a 0 = 2 2 c ε 1 2 π + 3 16 επa 2 = π 2 c 1 ε 1 / 2 + 3 8 ε 1 / 2 a 2 as ε 0.
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1 : Second-order differential equations in the phase plane 41 1.29 A mass m is attached to the mid-point of an elastic string of length 2 a and stiffness λ (see Figure 1.35 in NODE or Figure 1.44). There is no gravity acting, and the tension is zero in the equilibrium position. Obtain the equation of motion for transverse oscillations and sketch the phase paths. 1.29. Assume that oscillations occur in the direction of x (see Figure 1.44). By symmetry we can assume that the tensions in the strings on either side of m are both given by T . The equation of motion for m is 2 T sin θ = − m ¨ x . Assuming Hooke’s law, T = λ × extension = λ [ (x 2 + a 2 ) a ] . Therefore m ¨ x = − 2 kx [ (x 2 + a 2 ) a ] (x 2 + a 2 ) . ( i ) There is one expected equilibrium point at x = 0. This is a conservative system with potential (see NODE, Section 1.3) V (x) = 2 k x ax (x 2 + a 2 ) d x = k [ x 2 a (x 2 + a 2 ) ] . ( ii ) The equation of motion (i) can be expressed in the dimensionless form X = − X (X 2 + 1 ) 1 (X 2 + 1 ) after putting x = aX and t = mτ/( 2 k) . The phase diagram in the plane (X , Y = X ) is shown in Figure 1.45. From (ii) the potential energy V has a minimum at x = 0 (or X = 0) so that the origin is a centre. x m T T u Figure 1.44 Problem 1.29. 2 1 1 2 X Y 1 1 Figure 1.45 Problem 1.29: Phase diagram.
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42 Nonlinear ordinary differential equations: problems and solutions 1.30 The system ¨ x + x = F(v 0 − ˙ x) is subject to the friction law F(u) = 1 u > ε u/ε ε < u < ε 1 u < ε where u = v 0 − ˙ x is the slip velocity and v 0 > ε > 0. Find explicit equations for the phase paths in the (x , y = ˙ x) plane. Compute a phase diagram for ε = 0.2, v 0 = 1 (say). Explain using the phase diagram that the equilibrium point at ( 1, 0 ) is a centre, and that all paths which start outside the circle (x 1 ) 2 + y 2 = (v 0 ε) 2 eventually approach this circle. 1.30. The equation of the friction problem is ¨ x + x = F(v 0 − ˙ x) , where F(u) = 1 u > ε u/ε ε < u < ε 1 u < ε . The complete phase diagram is a combination of phase diagrams matched along the lines y = v 0 + ε and y = v 0 ε . y > v 0 + ε . In this region F = 1. Hence the phase paths satisfy d y d x = − x + 1 y , which has the general solution (x + 1 ) 2 + y 2 = C 1 . The phase paths are arcs of circles centred at ( 1, 0 ) . v 0 ε < y < v 0 + ε . The differential equation is ¨ x + x = 1 ε (v 0 − ˙ x) , or ε ¨ x + ˙ x + εx = v 0 , which is an equation of linear damping. The characteristic equation is εm 2 + m + ε = 0, which has the solutions m 1 , m 2 = 1 2 ε [− 1 ± ( 1 4 ε 2 ) ] .
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1 : Second-order differential equations in the phase plane 43 3 2 1 1 2 3 x 2 1 2 y y = v 0 + y = v 0 1 Figure 1.46 Problem 1.30: Phase diagram of ¨ x + x = F(v 0 − ˙ x) for ε = 0.2, v 0 = 1.
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