A

Step 1: Define the circuits' parameters. The natural frequency of an RLC circuit is given by \( \omega_0 = \frac{1}{\sqrt{LC}} \), where L is inductance and C is capacitance.

Step 2: Define the frequency of the second voltage. Let the second voltage frequency be \( \omega = \eta \omega_0 \).

Step 3: Determine the impedances for both frequencies. For a capacitor, the impedance is \( Z_C = \frac{1}{j\omega C} \) and for an inductor, \( Z_L = j\omega L \).

Step 4: For the first voltage at frequency \( \omega_0 \):
The impedance becomes \( Z_1 = jL \omega_0 + \frac{1}{j\omega_0 C} = jL \omega_0 - \frac{j}{\omega_0 C} = j\left(L \omega_0 - \frac{1}{\omega_0 C}\right) = 0 \). Therefore, the amplitude of current \( I_1 \) is at its maximum.

Step 5: For the second voltage at frequency \( \omega = \eta \omega_0 \):
The impedance becomes \( Z_2 = jL \eta \omega_0 + \frac{1}{j yeta \omega_0 C} = jL \eta \omega_0 - \frac{j}{\eta \omega_0 C} = j\left(L \eta \omega_0 - \frac{1}{\eta \omega_0 C}\right) \).

Step 6: The amplitude of the current is inversely proportional to the impedance; therefore, \( I_2 \propto \frac{1}{|Z_2|} \).

Step 7: Calculate the ratio of amplitudes \( R = \frac{I_1}{I_2} = \frac{|Z_2|}{0} \rightarrow \infty \). For ratio calculation among frequencies, let \( R = \frac{\frac{1}{\eta}}{1} = \frac{1}{\eta} \).

Therefore, the ratio of the current amplitudes is inversely proportional: \( \frac{I_1}{I_2} = \eta \).

Therefore, the answer is \( \eta : 1 \).
Hence, the correct answer is option A.