Exercises 5.4(c)To find critical points for the system in (a), we solvev= 0,−x+1λ−x= 0⇒v= 0,x2−λx+ 1 = 0⇒v= 0,x=λ±√λ2−42.For 0< λ <2,λ2−4<0 and so both roots are complex numbers. However, forλ >2there are two distinct real solutions,x1=λ−√λ2−42andx2=λ+√λ2−42,and the critical points areλ−√λ2−42,0andλ+√λ2−42,0.(d)The phase plane diagrams forλ= 1 andλ= 3 are shown in Figures B.40 and B.41 inthe answers section of the text.(e)From Figures B.40 we conclude that, forλ= 1, all solution curves approach the verticallinex= 1(=λ). This means that the bar is attracted to the magnet. The caseλ= 3is more complicated. The behavior of the bar depends on the initial displacementx(0)and the initial velocityv(0) =x(0). From Figure B.41 we see that (withv(0) = 0) if
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