The question appears to be "How do DC Capacitors get UL recognized?"
Or more generally "How does a product get UL Recognized or Listed?"
If the product fall under existing standards (UL and non UL standards) then the manufacturer applies to UL.
If the product does not fit the above standards or a change to the standard is required then the manufacture contacts the relevant standards committee representative (this could be UL ). For example ISO (http://www.iso.ch) or ANSI (http://www.ansi.org).
(The committees are composed of “interested parties” like UL, consumer groups, manufacturer’s, insurance representatives etc. )
Another route is to participate as a member of the UL Standards Technical Panel, I believe it takes substantial time and effort.
But if you want to participate try here ...
http://www.ul.com/global/eng/pages/solutions/standards/developstandards/participation/callformembers/index.jsp
The OP wanted specifically the history of how how the standards were developed for the capacitors.
The work of the committees / panels is recorded in minutes / bulitins however I don't know how to get hold of these.
Is amplitude of the waves modulated by the amount of current?
The amplitude of the current signal is half of the difference between the maximum and minimum of the current in each cycle of the oscillation.
For example, if the current over time is described by the equation
$$i(t) = A \sin(\omega{}t+\phi)$$
then A is the amplitude of signal because the current oscillates between +A and -A (and A has units of amps).
In the circuit shown, there is nothing modulating the current. Modulation happens when some other signal (like an audio waveform) changes the amplitude (for example) of the current waveform. This would require a more complex circuit than what is shown in your example.
Frequency is modulated by the frequency of the capacitor release of energy, correct?
Again, your circuit shows no mechanism for modulating the frequency. If you used a variable capacitor and the capacitance was controlled by another signal, that would cause frequency modulation. But it would probably also cause some undesirable parasitic amplitude modulation. To see some typical frequency modulator circuits, you can do a google image search for "frequency modulator" and look at the schematics that are found.
That would mean that using a capacitor with a lower farad rating should produce a higher frequency?
Generally the resonance of an LC tank circuit like in your example is given by
$$\omega_0=\frac{1}{\sqrt{LC}}$$
So the oscillation frequency can be increased by reducing either the capacitance or inductance value.
How would one go about keeping a sustained current to the inductor since it's constantly losing current without messing up the cycle?
You can use an oscillator circuit. Generally this means adding some kind of amplifier to the circuit to add energy to the waveform as quickly as it is lost to parasitic resistance and to feeding the load.
Best Answer
Guesstimate the resistances you would have in the real circuit, and insert resistors in the sim circuit to represent them.
A little more explanation of why the simulator reports an error without them may be useful : the instantaneous current when the battery and capacitor are connected is independent of the capacitance; it is purely the voltage divided by the resistances in series. Which, with ideal components, is the well known "divide by zero " error.
The capacitance will have some value of "ESR" = equivalent series resistance, which you may find from datasheets. (Some datasheets model ESR as loss, or "tan delta" instead) This page allows you to select capacitors by ESR; values from 0.003 ohms to 1.6 kilohm with 0.015 to a few ohms being common.
The battery will also have an ESR which will increase as the battery drains; again, see the datasheet. A fully charged NiCd or NiMh may have 0.1 ohm or more ESR, up to tens of ohms for a small button cell.
For the wiring or PCB traces, you can find the resistance in ohms per meter for approximately the right wire, (e.g. 0.1 ohm/m for 26SWG) and estimate the length of wire required (say 10cm, or 0.01 ohm).
Any one of these will break the loop and produce a circuit that can be simulated; modelling all of them will produce a more accurate simulation (may not change the results very much)
There may also be better simulation models for components such as capacitors, that allow you to set the ESR (and other parameters) and other features of real components.