17.5: Three Coupled Pendulums
( \newcommand{\kernel}{\mathrm{null}\,}\)
Let’s now move on to the case of three equal mass coupled pendulums, the middle one connected to the other two, but they’re not connected to each other.
The Lagrangian is
L=12mℓ2˙θ21+12mℓ2˙θ22+12mℓ2˙θ23−12mgℓθ21−12mgℓθ22−12mgℓθ23−12C(θ1−θ2)2−12C(θ3−θ2)2
Putting ω20=g/ℓ,k=C/mℓ2
L=12˙θ21+12˙θ22+12˙θ23−12ω20θ21−12ω20θ22−12ω20θ23−12k(θ1−θ2)2−12k(θ3−θ2)2
The equations of motion are
¨θ1=−ω20θ1−k(θ1−θ2)¨θ2=−ω20θ2−k(θ2−θ1)−k(θ2−θ3)¨θ3=−ω20θ3−k(θ3−θ2)
Putting θi(t)=Aieiωt, the equations can be written in matrix form
(ω20+k−k0−kω20+2k−k0−kω20+k)=ω20(100010001)+k(1−10−12−10−11)
The normal modes of oscillation are given by the eigenstates of that second matrix.
The one obvious normal mode is all the pendulums swinging together, at the original frequency ω0, so the springs stay at the rest length and play no role. For this mode, evidently the second matrix has a zero eigenvalue, and eigenvector (1,1,1).
The full eigenvalue equation is
|1−λ−10−12−λ−10−11−λ|=0
that is,
(1−λ)2(2−λ)−2(1−λ)=0=(1−λ)[(1−λ)(2−λ)−2]=(1−λ)(λ2−3λ)
so the eigenvalues are λ1=0,λ2=1,λ3=3, with frequencies
ω21=ω20,ω22=ω20+k,ω23=ω20+3k
The normal mode eigenvectors satisfy
(1−λ−10−12−λ−10−11−λ)(A1A2A3)=0
They are (1,1,1)/√3,(1,0,−1)/√2,(1,−2,1)/√6, normalizing them to unity.
The equations of motion are linear, so the general solution is a superposition of the normal modes:
(θ1θ2θ3)=1√3(111)Re(C1eiω1t)+1√2(10−1)Re(C2eiω2t)+1√6(1−21)Re(C3eiω3t)
