9780199212033

# If f mga there is one equilibrium position at θ 1 2

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If F 0 = mga there is one equilibrium position at θ = 1 2 π , in which the pendulum is horizontal. In this critical case the centre and the saddle merge at θ = 1 2 π so that the equilibrium point is a hybrid centre/saddle point. If F 0 > mga , there are no equilibrium positions. The phase diagram for = 1, g/a = 1 and F 0 /(ma 2 ) = 2 is shown in Figure 1.27. All phase paths approach the line ˙ θ = , which is a 2 3 2 2 1 1 2 A B 2 . Figure 1.26 Problem 1.14: Typical phase diagram for the friction-driven pendulum for F 0 < mga . 2 3 2 2 1 1 2 . 2 Figure 1.27 Problem 1.14: Typical phase diagram for the friction-driven pendulum for F 0 > mga .

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22 Nonlinear ordinary differential equations: problems and solutions ‘singular line’ along which the phase path continues. Whatever initial conditions are imparted to the pendulum, it will ultimately rotate at the same rate as the spindle. 1.15 By plotting ‘potential energy’ of the nonlinear conservative system ¨ x = x 4 x 2 , con- struct the phase diagram of the system. A particular path has the initial conditions x = 1 2 , ˙ x = 0 at t = 0. Is the subsequent motion periodic? 1.15. From NODE, (1.29), the potential function for the conservative system defined by ¨ x = x 4 x 2 is given by V (x) = − (x 4 x 2 ) d x = 1 3 x 3 1 5 x 5 . Its graph is shown in the upper diagram in Figure 1.28. The system has three equilibrium points: at x = 0 and x = ± 1. The equilibrium point at x = − 1 corresponds to a minimum of the potential function which generates a centre in the phase diagram, and there is a maximum at x = 1 which implies a saddle point. The origin is a point of inﬂection of V (x) . Near the origin ¨ x = − x 2 has a cusp in the phase plane. The two phase paths from the origin are given 1.5 1 0.5 0.5 1 1.5 x y 1 1 x 0.25 0.25 ( x ) 1 0.75 0.5 0.25 0.25 0.5 0.75 1 Figure 1.28 Problem 1.15: Potential energy and phase diagram for the conservative system ¨ x = x 4 x 2 .
1 : Second-order differential equations in the phase plane 23 by 1 2 y 2 = − 2 3 x 3 approximately and they only exist for x 0. Generally the equations for the phase paths can be found explicitly as 1 2 y 2 + V (x) = 1 2 y 2 + 1 3 x 3 1 5 x 5 = C . A selection of phase paths is shown in the lower diagram of Figure 1.28 including the path which starts at x( 0 ) = − 1 2 , y( 0 ) = 0. The closed phase path indicates periodic motion. 1.16 The system ¨ x + x = − F 0 sgn ( ˙ x) , F 0 > 0, has the initial conditions x = x 0 > 0, ˙ x = 0. Show that the phase path will spiral exactly n times before entering equilibrium (Section 1.6) if ( 4 n 1 )F 0 < x 0 < ( 4 n + 1 )F 0 . 1.16. The system is governed by the equation ¨ x + x = − F 0 sgn ( ˙ x) = F 0 ( ˙ x > 0 ) F 0 ( ˙ x < 0. For y > 0, the differential equation of the phase paths is d y d x = x F 0 y . Integrating, the solutions can be expressed as (x + F 0 ) 2 + y 2 = A . ( i ) Similarly, for y < 0, the phase paths are given by (x F 0 ) 2 + y 2 = B . ( ii ) For y > 0 the phase paths are semicircles centred at ( F 0 , 0 ) , and for y < 0 they are semicircles centred at (F 0 , 0 ) . The equation has a line of equilibrium points for which 1 < x < 1. The semicircle paths are matched as shown in Figure 1.29 (drawn with F 0 =

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