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I was reading a section of Introduction to Mechanics by Kleppner and Kolenkow:

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where it talks about the same time dependence. I'm not very familiar with this term but was wondering if there was some proof or intuition towards both pendulums having the same time dependence.

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    $\begingroup$ An important point is that such harmonic oscillators coupled linearly will not have the same time dependence in general. However, there are normal modes in which the oscillators do have the same harmonic time dependence, and it turns out that the general solution can always be expressed as a linear combination of the normal modes. $\endgroup$
    – Puk
    Mar 28 at 17:06

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If you interchange the two pendulums, you have the exact same system, so they better have the same time dependence.

Otherwise, it depends on which way you're looking at: from above the page, or from below the page...and that would be an unacceptable result.

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  • $\begingroup$ This answer makes no sense to me. The question is why the argument to $\sin$ must be the same in both equations, including the phase $\phi$. If $κ=0$ then the phases do not need to be the same, even though the system still has the interchange symmetry. $\endgroup$
    – benrg
    Mar 29 at 6:38
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Personally, I don't think this is intuitive at all. Every linear system has what is called normal modes$^\dagger$, solutions to the system where every component oscillates with the same frequency. A linear system is one where you can add two solutions together and still have a solution to the system.

So if we have such a system, there exist solutions where every part of the system oscillates with the same frequency. You can imagine a complicated setup of lots of springs and pendulums and somehow this has solutions where every part oscillates with the same frequency. This is quite amazing I think.

However, if you add two normal modes with different frequencies, this will break the pattern. There is not a single frequency anymore with which the system oscillates.

$\dagger$ I'm not entirely sure what the precise requirements are for a system having normal modes. I think it was that the system is linear. If someone can verify that it would be nice. This happens when every term in the equation is just $\theta(t)$ times a constant or a derivative of $\theta(t)$ times a constant.

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If you try slightly more general solutions

$$\begin{align} θ_1(t) = A_1\sin(ω_1 t + \phi_1) \\ θ_2(t) = A_2\sin(ω_2t+\phi_2) \end{align}$$

and plug them into Eqs. (6.5), you get

$$(Ω^2-ω_1²) A_1 \sin(ω_1 t + \phi_1) = κ^2 A_2 \sin(ω_2 t + \phi_2)$$

and its counterpart with 1 and 2 switched. If $κ\ne 0$, this equation says that the function $\sin(ω_1 t + \phi_1)$ is a constant multiple of the function $\sin(ω_2 t + \phi_2)$. That's only possible if $ω_1=ω_2$ and the phases differ by a multiple of 180°. The case where they are 180° out of phase can be handled by negating the amplitude instead, so you can take $\phi_1=\phi_2$.

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