Analysis Of Damped Pendulum Motion And Stability Analysis Of Its Equilibrium Points
Master pendulum motion with this comprehensive Homework Solution.
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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}.
d2θ/dt2 + 0.5 dθ/dt + 9.8 sinθ=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
2r2+r-19.6=0
r=-1+√(1+8.19.6)/2, -1-√(1+8.19.6)/2
let they are r1 and r2
Thus r2<0<r1
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}.
d2θ/dt2 + 0.5 dθ/dt + 9.8 sinθ=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
2r2+r-19.6=0
r=-1+√(1+8.19.6)/2, -1-√(1+8.19.6)/2
let they are r1 and r2
Thus r2<0<r1
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Subject
Physics