$$\boxed{\text{That circuit is not an oscillator}}$$

For it to act as a Hartley oscillator (as shown in this wiki page) then the inputs to the op-amp need to be swapped to get the correct phasing to make it sing at the resonant frequency of the LC circuit. The wiki page has had people indicate that there are problems with that particular circuit: –

So, if you re-arranged the inputs so that the centre tap of the two inductors fed the non-inverting input it would work more or less as Mr. Hartley intended. However, it won’t produce a very good sinewave because the op-amp open-loop gain will be quite high and the output waveform will crash into the power rails and give you somewhere between a full-on square wave and an unreliable and distorted sinewave.

So, if you wanted something like a decent op-amp based Hartley oscillator, I’d use this circuit: –

Transient analysis: –

It oscillates at 10.473 kHz in my simulator (micro-cap) and the formula above suggests it should run at 11.253 kHz. In other words, using a slow-ish op-amp creates delay in the feedback and it doesn’t quite sing “in tune”.

How I get the formula to calculate the frequency of this oscillator?

You derive it if you don’t trust what wiki tells you. Start with the parallel impedance of the two series inductors (L) and the capacitor (C): –

$$Z = \dfrac{j\omega L}{1 – \omega^2 LC}$$

Then, introduce the output impedance of the driver (R) and form a potential divider. It will have the following transfer function: –

$$\dfrac{j\omega L}{j\omega L + R – \omega^2 RLC}$$

Perform complex conjugate math to get a real term in the denominator: –

$$\dfrac{j(\omega LR – \omega^3 RL^2C) +\omega^2L}{\text{real number}}$$

And, we know that for the phase angle of the network to produce oscillations (when the correct polarity op-amp inputs are used), the imaginary parts in the above equation equate to zero and: –

$$\omega LR = \omega^3 RL^2C$$

Hence, the circuit will oscillate at: \$\hspace{1cm}\omega = \dfrac{1}{\sqrt{LC}}\$

Where \$L = L_1 + L_2\$.

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