Curves those of the pendulum equation for x 15 the

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curves those of the pendulum equation. For | x | < 1.5, the phase paths are visually indistin- guishable. The closed phase paths indicate periodic solutions, but the periods will increase with increasing amplitude. 1.12 The displacement, x , of a spring-mounted mass under the action of Coulomb dry friction is assumed to satisfy m ¨ x + cx = − F 0 sgn ( ˙ x) , where m , c and F 0 are positive constants (Section 1.6). The motion starts at t = 0, with x = x 0 > 3 F 0 /c and ˙ x = 0. Subsequently, whenever x = − α , where ( 2 F 0 /c) x 0 < α < 0 and ˙ x > 0, a trigger operates, to increase suddenly the forward velocity so that the kinetic energy increases by a constant amount E . Show that if E > 8 F 2 0 /c , a periodic motion exists, and show that the largest value of x in the periodic motion is equal to F 0 /c + E/( 4 F 0 ) . 1.12. The equation for Coulomb dry friction is m ¨ x + cx = − F 0 sgn ( ˙ x) = F 0 ˙ x > 0 F 0 ˙ x < 0 . For ˙ x = y < 0, the differential equation for the phase paths is given by m d y d x = F 0 cx y . Integrating this separable equation, we obtain 1 2 my 2 = − 1 2 c (F 0 cx) 2 + B 1 = 1 2 c (F 0 cx 0 ) 2 1 2 c (F 0 cx) 2 ,
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18 Nonlinear ordinary differential equations: problems and solutions C 1 C 2 y C 3 x x 0 x 1 a Figure 1.24 Problem 1.12: using the initial conditions x( 0 ) = x 0 , y( 0 ) = 0. This segment of the path is denoted in Figure 1.24 by C 1 . It meets the x axis again where F 0 cx = − F 0 + cx 0 , so that x = x 1 = 2 F 0 c x 0 . For y > 0, phase paths are given by 1 2 my 2 = − 1 2 c (F 0 + cx) 2 + B 2 . ( i ) Denote this segment by C 2 . It is the continuation into y > 0 of C 1 from x = x 1 , y = 0. Hence B 2 = 1 2 c (F 0 + cx 1 ) 2 = 1 2 c ( 3 F 0 x 0 ) 2 , so that x 1 = 2 F 0 c x 0 . The condition x 1 = − x 0 + ( 2 F 0 /c) < α ensures that the ‘trigger’ operates within the range of x illustrated. Denote the segment which meets C 1 at x = x 0 by C 3 . From (i), its equation is 1 2 my 2 = − 1 2 c (F 0 + cx) 2 + B 3 = − 1 2 c (F 0 + cx) 2 + 1 2 c (F 0 + cx 0 ) 2 . At x = − α , the energy on C 2 is E 2 = 1 2 c [ ( 3 F 0 cx 0 ) 2 (F 0 cα) 2 ] , whilst on C 3 , the energy is E 3 = 1 2 c [ (F 0 + cx 0 ) 2 (F 0 cα) 2 ] .
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1 : Second-order differential equations in the phase plane 19 At x = − α , the energy increases by E . Therefore E = E 3 E 2 = 1 2 c [ (F 0 + cx 0 ) 2 (F 0 cα) 2 ] − 1 2 c [ ( 3 F 0 cx 0 ) 2 (F 0 cα) 2 ] = 1 2 c [ (F 0 + cx 0 ) 2 ( 3 F 0 cx 0 ) 2 ] = 1 2 c [− 8 F 2 0 + 8 F 0 cx 0 ] A periodic solution occurs if the initial displacement is x 0 = E 4 F 0 + F 0 c . Note that the results are independent of α . For a cycle to be possible, we must have x 0 > 3 F 0 /c . Therefore E and F 0 must satisfy the inequality E 4 F 0 + F 0 c > 3 F 0 c , or E > 8 F 2 0 c . 1.13 In Problem 1.12, suppose that the energy is increased by E at x = − α for both ˙ x < 0 and ˙ x > 0; that is, there are two injections of energy per cycle. Show that periodic motion is possible if E > 6 F 2 0 /c , and find the amplitude of the oscillation. 1.13. Refer to the previous problem for the equations of the phase paths in y > 0 and y < 0.
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