Qualitative treatment:
The electromagnetic oscillations of LC system can be compared with the mechanical oscillations of a spring-mass system.
There are two forms of energy involved in LC oscillations. One is electrical energy of the charged capacitor, the other magnetic energy of the inductor carrying current.
Table: Energy in two oscillatory systems:

Likewise, the mechanical energy of the spring-mass system exists in two forms; the potential energy of the compressed or extended spring and the kinetic energy of the mass. The Table lists these two pairs of energy.
By examining the table, the analogies between the various quantities can be understood and these correspondences are given in the Table.
The angular frequency of oscillations of a spring-mass is given by,
\(\omega= \sqrt \frac{k}{m}\)
From Table, k→ 1/C and m → L. Therefore, the angular frequency of LC oscillations is given by,
\(\omega= \frac{1} {\sqrt {LC}}\)

During LC oscillations in LC circuits, the energy of the system oscillates between the electric field of the capacitor and the magnetic field of the inductor. Although, these two forms of energy vary with time, the total energy remains constant
It means that LC oscillations take place in accordance with the law of conservation of energy.
Total energy, \(U=U_E+U_B=\frac{q^{\prime}}{2 C}+\frac{1}{2} Li^2\)
Let us consider 3 different stages of LC oscillations and calculate the total energy of the system.
Case (i): When the charge in the capacitor, q = Q_mand the current through the inductor, i = 0, the total energy is given by,
\(U=\frac{Q_m^2}{2 C}+0=\frac{Q_{-}^2}{2 C}\)    ....(1)
The total energy is wholly electrical.
Case (ii): When charge = 0, current = I_m, the total energy is,
\(U =0+\frac{1}{2} L I_m^2=\frac{1}{2} L I_m^2  \)
\( =\frac{L}{2} \times\left(\frac{Q_m^2}{L C}\right)=\frac{Q_m^2}{2 C} \quad \text { since } I_m=Q_m \omega=\frac{Q_m}{\sqrt{L C}}\)   ...(2)
The total energy is wholly magnetic.
Case (iii): When charge = q, current = i, the total energy is,
\(U=\frac{q^2}{2 C}+\frac{1}{2} L i^2\)
Since, \(q=Q_m \cos \omega t, i=-\frac{d q}{d t}=Q_m \omega \sin \omega t.\)
The negative sign in current indicates that the charge in the capacitor decreases with time.
\(U=\frac{Q_{m}^2 \cos ^2 \omega t}{2 C}+\frac{L_\omega^2 Q_m^2 \sin ^2 \omega t}{2}\)
\(=\frac{Q_{m}^2 \cos ^2 \omega t}{2 C}+\frac{L Q_m^2 \sin ^2 \omega t}{2} \quad [since \omega^2=\frac{1}{LC}]\)
\(=\frac{Q_{m}^2}{2C} (cos ^2 \omega t+ sin^2 \omega t)\)
\(U=\frac{Q_{m}^2}{2C}\)   .....(3)
From above three cases, it is clear that the total energy of the system remains constant.

Inductive reactance is defined as the resistance offered by the inductors. Its unit is ohm.
X_L= ωL = 2πvL.
Capacitive reactance is defined as the resistance offered by the capacitors. Its unit is ohm.
\(X_c=\frac{1}{ωC}\)

(i) Consider a circuit containing a pure inductor of inductance L connected across an alternating voltage source (Figure). The alternating voltage is given by the equation.
v = V_m sin ωt .......(1)
(ii) The alternating current flowing through the inductor induces a self-induced emf or back emf in the circuit. The back emf is given by
Back emf, ε = \(-L\frac{di}{dt}\)
By applying Kirchoff's loop rule to the purely inductive circuit, we get
v + ε =0
V_m sin ωt = L \(\frac{di}{dt}\)
di = \(\frac{V_m}{L}\) sin ωt dt
Integrating both sides, we get
i = \(\frac{V_m}{L}\) ഽ sin ωt dt
i = \(\frac{V_m}{L_ω}\) (-cos ωt) + constant

(iii) The integration constant in the above equation is independent of time. Since the voltage in the circuit has only time dependent part, we can set the time independent part in the current (integration constant) into zero.
\(i=\frac { { V }_{ m } }{ \omega L } { sin }\left( \omega t-\frac { \pi }{ 2 } \right) \)   \(\left[ -{ cos\omega t=-sin\left( \frac { \pi }{ 2 } -\omega t \right) }\\ \because =sin\left( \omega t-\frac { \pi }{ 2 } \right) \right] \)
(or) \(i={ I }_{ m }sin\left( \omega t-\frac { \pi }{ 2 } \right) \) .....(2)
(iv) Where \(\frac { { V }_{ m } }{ \omega L } \) = I_m the peak value of the alternating current in the circuit. From equation (1) and (2), it is evident that - current lags behind the applied voltage \(\frac { \pi }{ 2 } \) in an inductive circuit. This fact is depicted in the phasor diagram. In the wave diagram also, it is seen that current lags the voltage by 90° .
(v) Inductive reactance X_L:
The peak value of current I_m is given by I_m = \(\frac { { V }_{ m } }{ \omega L } \). Let us compare this equation with I_m = \(\frac { { V }_{ m } }{ R} \) from resistive circuit The quantity ωL plays the same role as the resistance in resistive circuit. This is the resistance offered by the inductor, called inductive reactance (X_L) It is measured in ohm.
X_L = ωL