Current on Infinite Periodic LC Circuit

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The discussion revolves around deriving the equation of motion for an infinite periodic LC circuit and its relation to an infinite spring-mass system. The user is attempting to match the motion equations, specifically looking for a way to define the equilibrium length (A) in the context of the LC circuit. They seek guidance on transforming their current current function In(t) into a form that aligns with I(nA,t). Additionally, there is a request for clarification on symbols and the definition of "infinite periodic LC circuit." The user indicates they may have resolved their issue and inquires about closing the thread.
Kyuubi
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Homework Statement
Show that the current on an infinite periodic LC circuit obeys the wave equation in the long wavelength limit with the speed of the wave being the speed of light.
Relevant Equations
Equations of motion of current in an LC circuit.

(In)''=1/LC(-2In +In+1 +In -1)

Note here In means i sub n. As in the current on the nth inductor.
I wrote down the equation of motion for In(t) and I'm trying to match it with infinite spring mass system equation solution. In the spring mass system, we consider A to be the equilibrium length of the springs, and we can thus write Xn(t) = X(nA,t) and put it back into the equation of motion while taking a Taylor Expansion. This allows us to model the system as a long thin rod that's been pushed. But in the infinite periodic LC Circuit, what exactly is my A? What will help me turn my In(t) to I(nA,t)? In other words, how can I change my equation of motion
 
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Sorry I'm new to this. Is there a way to delete a thread or close it? I think I've solved my problem.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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