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I am dealing with a string-coupled pendulum, where two pendulum are tied onto one string as seen in image 1. (Image attributed to Young-ki Cho available from pre-view at DeepDyve)

coupled pendulum[1]

The symmetrical mode of the system entails that both pendulum are in phase and moving in simple harmonic motion. The equation for this mode is commonly known to be $\omega_s=\sqrt{g/l_s},$

Because the movement is simple, I believe that the system during the symmetrical mode can thus be visualized to be of natural frequency $f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}.$

I want to confirm what mode means. I have researched and looked at this definition: "A mode is a trajectory of a physical system which does not change shape as the system evolves. In other words, when a system is moving in a single mode, the positions of its parts all move with same general time dependence (e.g. sinusoidal motion with a single frequency) but may have different relative amplitudes."

Given that trajectory of this mode will not change shape, and that the position of all of its parts move with the same general time dependence, can I say that the mode can be represented by the wave velocity equation which is c = square root of g/l? It seems reasonable as "general time dependence" is the same for constant wave velocity.

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The time evolution of a system of coupled oscillators is, in general, described by a system of coupled second order differential equations. Its solution is not simple at all unless we are able to decouple them. By an appropriate choice of (generalized) coordinates, we are able to decouple the system of differential equations and the solutions for the coordinates are given as simple harmonic motions (SHM) of different frequencies, amplitudes and phases. Each of these SHM is called a mode and the complete solution of the system is a linear combination of these modes.

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  • $\begingroup$ I am aware that the motion can be represented by the superposition of this mode and one other, however, at a high-school level I cannot derive the modes. From your answer, am I correct in thinking that it is sensible to claim that the wave velocity equation at that SHM is equal to the mode? Thank you. $\endgroup$ – Peaceful Qualities Mar 17 '18 at 1:03
  • $\begingroup$ @GunaPrashant the motion of that SHM itself is the mode. Each mode is written as $A_n\sin(\omega_nt+\phi_n)$. Your original interpretation is quite good. $\endgroup$ – Diracology Mar 17 '18 at 1:07

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