A transistor -oscillator using a resonant circuit with an inductor `L` (of negligible resistance) and a capacitor `C` in series produce oscillations of frequency `f`. If `L` is doubled and `C` is changed to `4C`, the frequency will be

Options

A. `(f)/(4)`
B. `8f`
C. `(f)/(2sqrt(2))`
D. `(f)/(2)`

Answer

To solve the problem, we need to analyze how the frequency of oscillation changes when the values of the inductor \( L \) and the capacitor \( C \) are modified.

### Step-by-Step Solution:

1. **Understand the formula for frequency in an LC circuit**:
The frequency \( f \) of oscillation in a resonant LC circuit is given by the formula:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]

2. **Identify the initial parameters**:
Let the initial values of the inductor and capacitor be:
- Inductor: \( L \)
- Capacitor: \( C \)
Thus, the initial frequency \( f \) can be expressed as:
\[ f = \frac{1}{2\pi \sqrt{LC}} \]

3. **Modify the parameters as per the problem statement**:
According to the question:
- The inductor is doubled: \( L' = 2L \)
- The capacitor is changed to four times its original value: \( C' = 4C \)

4. **Substitute the new values into the frequency formula**:
The new frequency \( f' \) with the modified values is:
\[ f' = \frac{1}{2\pi \sqrt{L'C'}} = \frac{1}{2\pi \sqrt{(2L)(4C)}} \]

5. **Simplify the expression**:
Now simplify the expression for \( f' \):
\[ f' = \frac{1}{2\pi \sqrt{8LC}} = \frac{1}{2\sqrt{2} \cdot 2\pi \sqrt{LC}} = \frac{1}{2\sqrt{2}} \cdot \frac{1}{2\pi \sqrt{LC}} \]
Since \( \frac{1}{2\pi \sqrt{LC}} = f \), we can write:
\[ f' = \frac{f}{2\sqrt{2}} \]

6. **Conclusion**:
Therefore, the new frequency \( f' \) after modifying the inductor and capacitor is:
\[ f' = \frac{f}{2\sqrt{2}} \]

### Final Answer:
The frequency will be \( \frac{f}{2\sqrt{2}} \).