I have obtained the following equations of motion $$2\ddot \theta_1 +\ddot \theta_2 \cos(\theta_2-\theta_1) -(\dot \theta_2)^2 \sin(\theta_2-\theta_1)+\frac{2g}{l}\sin(\theta_1)=0$$ and $$\frac{2}{3}\ddot \theta_2 +\ddot \theta_1 \cos(\theta_2-\theta_1) +(\dot \theta_1)^2\sin(\theta_2-\theta_1)+\frac{g}{l}\sin\theta_2=0$$

I then proceeded to linearise them and got the following equations

$$2\ddot \theta_1 +\ddot \theta_2+\frac{2g}{l}\theta_1=0$$

and $$\frac{2}{3}\ddot \theta_2 +\ddot \theta_1+\frac{g}{l}\theta_2=0$$ for the following problem enter image description here

i was wondering if anyone could show me a solution for 6, because i'm stuck and its the last part, i know we need to find the eigenvalues but i'm terrible at writing the equations in matrices.

  • $\begingroup$ i'm really stuck on this problem, $\endgroup$ – user395952 Apr 5 '17 at 18:10

You replace

  • $\cos(θ_2−θ_1)$ with $1$
  • $(\dot θ_k)^2\sin(θ_2−θ_1)$ and similar higher order terms with $0$
  • $\sinθ_k$ with $θ_k$.

and should obtain greatly simplified equations. See also "double-pendulum", as you could replace the bar with two point-masses at the ends. There should be some posts here and elsewhere on that topic.

In the end you will get a system $A\ddot θ+Bθ=0$. You could treat this as first order system of dimension $4$, $$ \pmatrix{\ddot θ\\\dot θ} = \pmatrix{0&-A^{-1}B\\I&0} \pmatrix{\dot θ\\θ} $$ and solve this via eigen decomposition, or you could directly insert $θ=θ_0e^{i\omega t}$ to get the eigenvalue problem $$ (Aω^2-B)θ_0=0\iff (A^{-1}B)θ_0=ω^2θ_0 $$ where you can find the eigenvalues from $\det(Aω^2-B)=0$ without inverting matrices.

  • $\begingroup$ what about $\ddot \theta$ ? or does that stay the same? $\endgroup$ – user395952 Apr 3 '17 at 14:54
  • $\begingroup$ Yes, they stay as they are first order terms. You only remove quadratic and higher order terms. $\endgroup$ – Lutz Lehmann Apr 3 '17 at 16:34
  • $\begingroup$ so is this correct $$2\ddot \theta_1 +\ddot \theta_2+\frac{2g}{l}\theta_1=0$$ for one of the equations of motion? $\endgroup$ – user395952 Apr 3 '17 at 19:09
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Lutz Lehmann Apr 3 '17 at 19:26
  • $\begingroup$ okay so now i've linearised them how do i go about finding the natural frequencies and modes $\endgroup$ – user395952 Apr 3 '17 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy