I am studying an RLC circuit with a switch which initial position is \$a\$ as stated in the following image and with values \$R_g=5000\Omega\$, \$L_g=320H\$, \$C_1=31.66nF\$ and \$V_g=311\cdot\sin(100\pi t)\$:
The switch goes from position \$a\$ to position \$b\$ at \$t=0.41s\$, when \${v_c}_1(t)\$ reaches \$6000kV\$. The expression of the voltage across capacitor \$C_1\$ is the following one:
$$ {v_C}_1(t)=-6260\cdot\cos(314t)+11.4\cdot\sin(314t)+e^{-7.81t}\cdot(6260\cdot\cos(314t)+144\cdot\sin(314t))$$
Using PSpice to plot the voltage across the capacitor \$C_1\$ gives me the following graph (it reaches Sinusoidal Steady State):
The Laplace equivalent circuit when switch is in position \$b\$ is the following:
Switch \${S_w}_2\$ is not important in this question. The expression of the voltage in the capacitor \$C_2\$ after computing it is the following:
$${{{v_S}_w}_2}\left(t\right)=117440\cdot\left(\cos\left(9.71\cdot10^5t\right)-\cos\left(1.09\cdot10^6t\right)\right)$$
Plotting voltages across \$C_1\$ and \$C_2\$ when \$t>0.41s\$ gives me the following graph (it is zoomed in a lot):
My question is, how is it possible that the frequency of the voltage that provides \$C_1\$ (since in circuit with the switch in position \$b\$, \$C_1\$ acts as generator) increases that way? Is it because of the coupled inductors? Why is it not the frequency of \$V_g\$?