Peak Voltage

The peak voltage between $A$ and $B$ is given by:

$\begin{aligned} V_{AB} & = V_A - V_B \\ & = \frac {\frac 1{j\omega C}}{j\omega L+\frac 1{j \omega C}}V_s - \frac {j \omega L}{j\omega L+\frac 1{j \omega C}}V_s \\ & = \frac {\frac 1{j\omega C}-j\omega L}{j\omega L+\frac 1{j \omega C}}V_s & \small \blue{\text{Multiply up and down by }j \omega C} \\ & = \frac {1+\omega^2 LC}{1-\omega^2 LC} V_s \\ \implies \alpha & = \frac {1+\omega^2 LC}{1-\omega^2 LC} \\ \alpha - \alpha \omega^2LC & = 1 + \omega^2 LC \\ \implies \omega & = \sqrt{\frac {\alpha -1}{(\alpha +1)LC}} \end{aligned}$

Therefore, $\beta+\gamma = 1 - 1 = \boxed 0$ .

Let the angular frequency be $\omega$ . The capacitive reactance and inductive reactance is:

$X_C = -\frac{j}{\omega C}$ $X_L = j\omega L$

Let the source current be $I$ , the current flowing through $A$ be $I_A$ and that through $B$ be $I_B$ . The branches $A$ and $B$ are in parallel. Therefore, the equivalent impedance of the given circuit across the source ends is:

$Z = \frac{X_L+X_C}{2}$

Therefore, the current flowing through the source is:

$I = \frac{V_S}{Z}$ $\implies I = \frac{2V_S}{X_L+X_C}$

Now, since the branches $A$ and $B$ are similar, one can conclude (and even prove) that:

$I_A = I_B = 0.5I= \frac{V_S}{X_L+X_C}$

Let the negative terminal of the source be grounded (at $0 \ V$ ). Then:

$V_S - V_A = I_AX_L$ $V_S - V_B = I_BX_C$ $\implies V_A - V_B = \Delta V = \frac{V_S(X_C-X_L)}{X_C+X_L}$

Plugging in all expressions and computing the modulus of $\Delta V$ gives:

$\lvert \Delta V \rvert = \frac{V_S(\omega^2LC+1)}{\omega^2LC-1}$

Using the fact that $\lvert \Delta V \rvert =\alpha V_S$ and solving for $\omega$ gives:

$\omega^2 = \frac{\alpha+1}{(\alpha-1)LC}$