Electronic – Does a voice coil vibrate at the same frequency as the source sound file

Gabric - here are my answers to the 4 questions:

1.) The oscillation starts as soon as you close the switch. This closing causes a sharp inrush transient which contains (among others) also the frequency component that fulfills Barkhausen`s oscillation condition (see point 2).

2.) The oscillation condition requires a loop gain (slighly) larger than unity and a loop phase of -360 deg (0 deg). The transistor in common emitter configuration contributes -180 deg phase shift between base and collector. The remaining phase shift is provided by the feedback loop.

This loop again consists of a third-order lowpass which is able to provide these -180 deg at one single frequency only. The lowpass consists of a second-order lowpass (L1-C) working upon a first-order lowpass (Lq-r,in). Here, Lq is the crystal working as an high-quality inductor and r,in is the finite input resistance of the transistor at the base node.

3.) The primary inductor is tapped to provide a certain kind of ac voltage division in order to limit the gain of the amplifier stage (because it should not be too large, just sufficient to allow self-excitement without severe non-linearities).

4.) The 10k resistor is necessary to provide the correct DC biasing of the transistor; the small emitter resistor provides dc feedback for stabilizing the dc bias point. The 0.05µF capacitor establishes signal ground potential at the pos. supply pin. This is necessary because the capacitor of the first L-C lowpass block is connected to the pos. supply but must be referenced to ac ground.

If the passive resonant circuit has a high 'Q' then energy can build up in the circuit over many cycles. It will similarly die down naturally over many cycles.

The intensity is not "ever increasing", however. As the intensity increases, so do the losses and at some point the losses equals the input power, so you have an equilibrium.

Imagine a tuning fork. If you keep exciting it, the vibrations will build up (as will the losses) and at some point the metal shape might change if the excitation is powerful enough, but it's unlikely. Most real circuits have a rather lower Q than the mechanical Q of a tuning fork.

In a real LC resonant circuit, the losses are usually due to inductor resistance, to core losses (if a core is used) and to electromagnetic radiation, especially at higher frequencies. Capacitor dielectric losses contribute too. Superconducting circuits can have huge Q's (in the thousands), resonant circuits made with parts from your favorite distributor, much more disappointing.