I've made a small illustration depicting the key idea. If this is in coherence with what you've asked, we could summarize some important points about the case.
Total energy of the Earth and Bar Magnet system is given by the equation: $KE + PE = \frac{1}{2}mv^{2} + \frac{GMm}{R}$
While PE is there for both Earth and magnet system (combined), KE is available for the magnet to spend on oscillation. If, while oscillating, it spends this KE by transforming it in to thermal energy in the resistor-inductor circuit, KE decreases slowly and becomes zero. $KE + PE = 0 + \frac{GMm}{R}$ ie, the oscillation seize to exist.
An easy way to look at it would be to consider change in KE. We will start with the magnetic pendulum:
Component of angular acceleration = $\alpha =\frac{-Lmg\ sin(\theta)cos(\theta)}{I}$
This clearly has its peak values at 45 degree amplitudes. 780
Now, component force due to this restoring torque is give by,
$F =\frac{-L^{2}m^{2}g\ sin(\theta)cos(\theta)}{I}$
From this pushing or pulling force we can derive the power and thereby energy:
$P=Fv$
$\frac{dE}{dt} = Fv = \frac{-L^{2}m^{2}gv}{I}sin(\theta)cos(\theta)$
On integrating,
$\int dE = \int \frac{-L^{2}m^{2}gv}{I}sin(\theta)cos(\theta)\ dt$
$\int_{0}^{\frac{T}{2}}dE = k\int_{0}^{\frac{T}{2}} vsin(\theta)cos(\theta)\ dt$
This gives the push energy in one single push (half the oscillation period).
Now the energy burned by the resistor is given by:
$dE = Pdt$
$\int_{0}^{\frac{T}{2}}dE = \int_{0}^{\frac{T}{2}}Pdt$
$\int_{0}^{\frac{T}{2}}dE = \int_{0}^{\frac{T}{2}}I^{2}Rdt$
Current in the LR circuit is
$I = \frac{V}{R}\left (1-e^{\frac{-Rt}{L}} \right )$
Substituting for current,
$\int_{0}^{\frac{T}{2}}dE = \int_{0}^{\frac{T}{2}} V\left (1-e^{\frac{-Rt}{L}} \right )^{2}dt $
Thus the change in energy for a half cycle is
$\Delta E_{\frac{T}{2}} = k\int_{0}^{\frac{T}{2}} vsin(\theta)cos(\theta)\ dt - \int_{0}^{\frac{T}{2}} V\left (1-e^{\frac{-Rt}{L}} \right )^{2}dt$
This is how the RHS dampens energy change. The resistance in the second integral dominates in each integration. If it wasn't there, The remaining emf/potential term will cancel itself in a full cycle integration. Thus, there would not be an energy change. This isn't surprising because inductor can give back the magnetic force in each half cycle with the same magnitude. I hope this will make it more clear since my previous attempt was somewhat dubious. 37