SOLUTION

An inductor’s impedance is twice that of a capacitor at a specific frequency (f).

Both components are connected in series.

We need to find the new resonant frequency.

Impedance relationship

The inductive reactance X_L and capacitive reactance X_C are related by:

X_L = 2X_C

Substituting their formulas:

$\omega L = 2 \cdot \frac{1}{\omega C}$

where

$ \omega = 2\pi f $ is the angular frequency.

Step 2: Solve for angular frequency $$ \omega $$

Rearranging the equation:

$\omega^2 L = \frac{2}{C}$

$\omega = \sqrt{\frac{2}{LC}}$

Step 3: Resonant frequency formula

The resonant angular frequency is related to the resonant frequency ($ f_\text{new}$) by:

$\omega = 2\pi f_\text{new}$

Substitute $ \omega = \sqrt{\frac{2}{LC}}$:

$2\pi f_\text{new} = \sqrt{\frac{2}{LC}}$

Solve for $ f_\text{new}$:

$f_\text{new} = \frac{\sqrt{2}}{2\pi \sqrt{LC}}$

Step 4: Compare with the original resonant frequency

The original resonant frequency ($ f_\text{original}$) is given by:

$f_\text{original} = \frac{1}{2\pi \sqrt{LC}}$

Now compare:

$f_\text{new} = \sqrt{2} \cdot f_\text{original}$

The new resonant frequency is:

$f_\text{new} = \sqrt{2}f$