In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes x times its initial resonant frequency `omega_0`. The value of x is.

Options

A. 45017
B. 4
C. 16
D. 42370

Answer

To solve the problem, we need to determine how the resonant frequency of an LC oscillator changes when the inductance and capacitance are altered.

### Step-by-Step Solution:

1. **Understand the formula for resonant frequency**:
The resonant frequency \( \omega_0 \) of an LC oscillator is given by the formula:
\[ \omega_0 = \frac{1}{\sqrt{LC}} \]
where \( L \) is the inductance and \( C \) is the capacitance.

2. **Identify the new values of inductance and capacitance**:
According to the problem, the inductance becomes twice its original value:
\[ L' = 2L \]
and the capacitance becomes eight times its original value:
\[ C' = 8C \]

3. **Calculate the new resonant frequency**:
The new resonant frequency \( \omega' \) can be expressed as:
\[ \omega' = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(2L)(8C)}} \]
Simplifying this gives:
\[ \omega' = \frac{1}{\sqrt{16LC}} = \frac{1}{4\sqrt{LC}} \]

4. **Relate the new frequency to the initial frequency**:
Now, we can express the ratio of the new frequency \( \omega' \) to the initial frequency \( \omega_0 \):
\[ \frac{\omega'}{\omega_0} = \frac{\frac{1}{4\sqrt{LC}}}{\frac{1}{\sqrt{LC}}} = \frac{1}{4} \]

5. **Determine the value of \( x \)**:
From the above ratio, we can see that:
\[ \omega' = \frac{1}{4} \omega_0 \]
Therefore, \( x \) is given by:
\[ x = \frac{1}{4} \]

### Final Answer:
The value of \( x \) is \( \frac{1}{4} \).