Double pendulum: normal modes and time evolution

Last time, I derived the linearised equations of motion of a double pendulum. This video shows how to solve them to find the normal modes and hence the general time evolution of the system.
Deriving the equations of motion: • Double pendulum: equat...
About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
My website: benyelverton.com/
#physics #mathematics #doublependulum #pendulum #dynamics #mechanics #normalmodes #calculus #matrices #vectors #linearalgebra #determinant #oscillations #smalloscillations #maths #math #science #education

[@uttamraj3786]

Great video sir keep uploading.

[@DrBenYelverton]

Thanks! Will do.

[@jyotimandal895]

Thank you, sir.

[@muhammadazamuddinbinrosly9150]

good video!!

[@DrBenYelverton]

Thanks for watching, I'm glad you enjoyed it!

[@LoGos7891]

At 10:15 you say that we can ignore the +/- factors because cosine is even. But the phase-shifted cosine function that we have assumed is NOT necessarily even. Have I missed something?

[@DrBenYelverton]

You're right that cos(ωt+φ) is not necessarily an even function of t, but it's still an even function of its argument as a whole. This means that cos(-ωt+φ) = cos(ωt-φ). You could then let ψ = -φ and write it as cos(ωt+ψ). Since the phase is arbitrary (to be determined by initial conditions), cos(ωt+φ) and cos(ωt+ψ) are completely equivalent solutions and the sign of ω therefore doesn't matter.