In a L-C circuit, angular frequency at resonance is `omega`. What will be the new angular frequency when inductor's inductance is made two times and capacitor's capacitance is made four times?

Options

A. `omega/2sqrt(2)`
B. `omega/sqrt(2)`
C. `2omega`
D. `(2omega)/(sqrt(2))`

Explanation

To solve the problem, we need to find the new angular frequency at resonance when the inductance and capacitance of an LC circuit are changed.

### Step-by-Step Solution:

1. **Understand the Resonant Frequency Formula**:
The angular frequency at resonance (ω_r) for an LC circuit is given by the formula:
\[ \omega_r = \frac{1}{\sqrt{LC}} \]
where L is the inductance and C is the capacitance.

2. **Identify the Changes in Inductance and Capacitance**:
According to the problem:
- The inductance is doubled: \( L' = 2L \)
- The capacitance is quadrupled: \( C' = 4C \)

3. **Substitute the New Values into the Resonant Frequency Formula**:
We need to find the new resonant frequency \( \omega_r' \) using the modified inductance and capacitance:
\[ \omega_r' = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(2L)(4C)}} \]

4. **Simplify the Expression**:
Now simplify the expression:
\[ \omega_r' = \frac{1}{\sqrt{8LC}} = \frac{1}{2\sqrt{2}\sqrt{LC}} \]

5. **Relate to the Original Resonant Frequency**:
Recall that \( \omega_r = \frac{1}{\sqrt{LC}} \). Therefore, we can express \( \omega_r' \) in terms of \( \omega_r \):
\[ \omega_r' = \frac{\omega_r}{2\sqrt{2}} \]

6. **Final Answer**:
Thus, the new angular frequency at resonance when the inductance is doubled and the capacitance is quadrupled is:
\[ \omega_r' = \frac{\omega}{2\sqrt{2}} \]