Oscillator problem

Here you will find a good explanation using a PNP transistor instead of a NPN, but the way it works is the same. In what follows I will describe the general idea.

To understand how the feedback occurs look at the picture below. The red U is the fork (for illustrative purpose). Each leg of the fork has a permanent magnet that interacts with the drive coils and the feedback coils. When it is oscillating around its equilibrium position, the maximum speed will be achieved when the legs are passing through that equilibrium position, like a mass on a spring. It is during maximum speed that the induction will be maximum.

Now suppose that the capacitor is initially discharged, then the voltage from Vdc will bring the base voltage above the emitter voltage and the transistor will turn on. The current that will flow through the drive coils D1 and D2 will be enough to start the oscillations. Meanwhile, the base current will start to charge the capacitor. In steady state the voltage induced in the feedback coil F1 will be turning the transistor on and off, with the capacitor barely discharging during the off cycles (C*R >> 1/(frequency of oscillations ).

Suppose the the fork is going through the point of equilibrium, then it will induce a voltage in F1, with the sign depending on the direction of movement. Only one of those peaks of voltage will turn the transistor on, the one which brings the base voltage above the emitter's. When that happens the current through D1 and D2 will sustain the fork oscillation by creating a magnetic field that 'pushes' the permanent magnets in the fork.

Next, suppose that the oscillations grow larger in amplitude and let's see how the circuit corrects it. First, a higher voltage will be induced in F1, but that shouldn't matter when the transistor is fully on (saturated). On the other hand, a higher voltage will also be induced in D1 and D2 and it will be such that it will tend to oppose the current drawn by the transistor through the collector. The result is that the driving current (collector current) decreases, and that means a smaller magnetic field and consequently a smaller "push" to the fork. Therefore the oscillation amplitude will decrease.

Another way by which the circuit corrects itself is through the capacitor C. If the oscillations get larger then a higher voltage will be induced in F1, but if the transistor is not fully saturated then a higher current will flow through the base in response to the higher voltage in the base with respect to the emitter. That current will charge C during the ON cycle, which will begin to discharge during the OFF cycle through R. So a higher base current means a higher voltage in the capacitor at the end of the ON cycle, which means that it will be harder for the next peak of voltage to overcome that new capacitor voltage and turn on the transistor in the next cycle. Remember the RC constant is relatively big, so the charge time of C is larger than the discharge time.

I think that's all. If you are wondering how a vibrating fork could replace the escapement of a mechanical watch watch this video.

It's not the same circuit on the emitter. You've got a 1k pot in series with a 1k resistor and the pot wiper connects to a grounding capacitor. You've also got a 10pF cap where it is 100nF in LTSpice.

I'm not saying any of these matter but you can't possibly make comparisons with these glaring disparities.

EDIT

Following comments from the OP, the problem I think arises from the new "intended" operating frequency of the circuit (he uses a 10pF cap on the multisim circuit to set the resonant frequency whilst the LTSpice circuit has 100nF). Added to this is the lack of emitter capacitor - the multisim circuit shows it connected via the wiper of a pot set midway and this will drastically reduce the gain and prevent or delay oscillation. Remember that the collector has to generate enough signal to drive the 10nF capacitor (C3) and at a much higher frequency this 10nF (via C5, 1uF) will probably look like a short circuit - the gain of the transistor has to be greater than 1 for it to begin oscillation and with no directly connected emitter capacitor and a 10nF effectively connected to the collector I think the gain will be less than unity.

I'm not totally familiar with this type of Colpitts oscillator but it seems to me that the LTSpice circuit should oscillate at about 41kHz whereas the MS circuit should operate at about 4.1MHz - this isn't going to happen with a 10pF - its reactance at 4.1MHz will be nearly 4 kohms and this will get "battered" by R1 and R2 in the imedance versus resistance race. Do the sensible thing and start with the same circuit. On the LT circuit, the 100nF will have an impedance of about 40 ohms at 41kHz and therefore not be hardly affected by R1 and R2.