Analysis of Damped Pendulum Motion and Stability Analysis of Its Equilibrium Points1.Given the equation of motion for a damped pendulum:d2θdt2+0.5dθdt+9.8sinθ=0\frac{d^2\theta}{dt^2} + 0.5\frac{d\theta}{dt} + 9.8\sin\theta= 0, whereθ\thetais the angle andy=dθdty =\frac{d\theta}{dt}, derive the first-order systemand express it in terms ofx=θx =\thetaandy=dθdty =\frac{d\theta}{dt}.d2θ/dt2+ 0.5 dθ/dt + 9.8 sinθ=0Here x=θ , y=dθ/dtFrom the above equation , we haved(y)/dt+ 0.5 (y)+9.8 sinθ=0putting the value for dθ/dt=ydy/dt+ 0.5 y+9.8 sinx=0dy/dt=-0.5 y-9.8 sinx …………….(i)x=θdifferentiate it w.r.t. t , we get dx/dt = dθ/dt=ydx/dt=y………….(ii)2)At the equilibrium point(π,0)(\pi, 0), analyze the stability of the system and find the eigenvaluesof the linearized system.A={(0,1),(-9.8cosx–0.5)}Hence,at equilibrium point(π,0)A={(0,1),(9.8,-0.5)}equation of eigenvalues are r(r+0.5)-9.8=02r2+r-19.6=0r=-1+√(1+8.19.6)/2,-1-√(1+8.19.6)/2letthey are r1and r2Thus r2<0<r1Preview Mode
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