Answer
The angular frequency of oscillation of the four-element circuit is $~~\omega$
Work Step by Step
It is given that $\omega = \frac{1}{\sqrt{L_1~C_1}} = \frac{1}{\sqrt{L_2~C_2}}$
Then:
$L_1~C_1 = L_2~C_2 = \frac{1}{\omega^2}$
The equivalent inductance of the four-element circuit is $L = L_1+L_2$
We can find the equivalent capacitance of the four-element circuit:
$\frac{1}{C} = \frac{1}{C_1}+\frac{1}{C_2}$
$\frac{1}{C} = \frac{C_2}{C_1~C_2}+\frac{C_1}{C_1~C_2}$
$C = \frac{C_1~C_2}{C_1+C_2}$
We can find the angular frequency of oscillation of the four-element circuit:
$\omega' = \frac{1}{\sqrt{L~C}}$
$\omega' = \frac{1}{\sqrt{(L_1+L_2)~(C_1~C_2)/(C_1+C_2)}}$
$\omega' = \frac{\sqrt{C_1+C_2}}{\sqrt{L_1~C_1~C_2+L_2~C_1~C_2}}$
$\omega' = \frac{\sqrt{C_1+C_2}}{\sqrt{C_2/\omega^2+C_1/\omega^2}}$
$\omega' = \frac{\omega~\sqrt{C_1+C_2}}{\sqrt{C_2+C_1}}$
$\omega' = \omega$
The angular frequency of oscillation of the four-element circuit is $~~\omega$
