132 take moments about the point of suspension of the

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1.32. Take moments about the point of suspension of the pendulum. Then Fh sin φ mga sin θ = ma 2 ¨ θ . ( i ) where, by the inverse square law, F = c a 2 + h 2 2 ah cos θ , tan φ = a sin θ h a cos θ . ( ii ) Elimination of f and φ in (i) using (ii) leads to an equation in θ ma ¨ θ = ch sin θ (a 2 + h 2 2 ah cos θ) 3 / 2 mg sin θ . There are equilibrium points at θ = , (n = 0, ± 1, ± 2, . . .) and where a 2 + h 2 2 ah cos θ = ch mg 2 / 3 , that is where cos θ = a 2 + h 2 (ch/mg) 2 / 3 2 ah . This equation has solutions if 1 a 2 + h 2 (ch/mg) 2 / 3 2 ah 1, or mg(a h) 3 h c mg(a + h) 3 h . ( iii ) If c lies outside this interval then the pendulum does not have an inclined equilibrium position. If it exists let the angle of the inclined equilibrium be θ = θ 1 for 0 < θ 1 < π . Obviously θ = − θ 1 and 2 ± θ 1 will also be solutions.
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46 Nonlinear ordinary differential equations: problems and solutions This is a conservative system with potential V (θ) (see Section 1.3) such that V (θ) = − ch sin θ ma(a 2 + h 2 2 ah cos θ) 3 / 2 + g sin θ a . The nature of the stationary points can be determined by the sign of the second derivative at each point. Thus V (θ) = − ch cos θ ma(a 2 + h 2 2 ah cos θ) 3 / 2 + 3 ch 2 sin 2 θ m(a 2 + h 2 2 ah cos θ) 5 / 2 + g cos θ a . θ = ( n even). V (nπ) = − ch ma(h a) 3 + g a . It follows that θ = is a centre if c < mg(h a) 3 /h and a saddle if c > mg(h a) 3 /h . θ = ( n odd). V (nπ) = ch ma(h + a) 3 g a . Therefore θ = is a centre if c > mg(h + a) 3 /h and a saddle if c < mg(h + a) 3 /h . θ = θ 1 subject to mg(a h) 3 ch mg(a + h) 3 . V 1 ) = − ch cos θ 1 ma(a 2 + h 2 2 ah cos θ 1 ) 3 / 2 + 3 ch 2 sin 2 θ 1 m(a 2 + h 2 2 ah cos θ 1 ) 5 / 2 + g cos θ 1 a = mg ch ( 5 / 3 ) ch 2 sin 2 θ 1 m > 0. Note that V ( θ 1 ) is also positive Therefore, if they exist, all inclined equilibrium points are centres. Suppose that the parameters a , h and m are fixed, and that c can be increased from zero. The behaviour of the bob is as follows: 0 < c < mg(a h) 3 /h . There are two equilibrium positions: the bob vertically below the suspension point which is a stable centre, or the bob above which is an unstable saddle, c takes the intermediate values defined by (iii). Both the highest and lowest points becomes a saddles. The inclined equilibrium points are centres. c > mg(a + h) 3 /h . The lowest point remains a saddle but the highest point switches back to a saddle.
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1 : Second-order differential equations in the phase plane 47 1.33 A pendulum with equation ¨ x + sin x = 0 oscillates with amplitude a . Show that its period, T , is equal to 4 K(β) , where β = sin 2 1 2 a and K(β) = 1 2 π 0 d φ ( 1 β sin 2 φ) . The function K(β) has the power series representation K(β) = 1 2 π 1 + 1 2 2 β + 1.3 2.4 2 β 2 + · · · , | β | < 1. Deduce that, for small amplitudes, T = 2 π 1 + 1 16 a 2 + 11 3072 a 4 + O(a 6 ) .
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