Analysis Of Damped Pendulum Motion And Stability Analysis Of Its Equilibrium Points

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Analysis of Damped Pendulum Motion and Stability Analysis of Its Equilibrium Points 1. 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 , where θ \ theta is the angle and y=dθdty = \ frac{d \ theta}{dt} , derive the first - order system and express it in terms of x=θx = \ theta and y=dθdty = \ frac{d \ theta}{dt} . d 2 θ /dt 2 + 0.5 d θ /dt + 9.8 sin θ =0 Here x=θ , y=dθ/dt From the above equation , we have d(y)/dt+ 0.5 (y)+9.8 sinθ=0 putting the value for dθ/dt=y dy/dt+ 0.5 y+9.8 sinx=0 dy/dt= - 0.5 y - 9.8 sinx …………….(i) x=θ differentiate it w.r.t. t , we get dx/dt = dθ/dt=y dx/dt=y ………….(ii) 2) At the equilibrium point (π,0)( \ pi, 0) , analyze the stability of the system and find the eigenvalues of 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=0 2r 2 +r - 19.6=0 r= - 1+√(1+8.19.6)/2, - 1 - √(1+8.19.6)/2 let they are r 1 and r 2 Thus r 2 <0<r 1
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