At a certain frequency `omega_(1)`, the reactance of a certain capacitor equals that of a certain inductor. If the frequency is changed to `omega_(2) = 2omega_(1)`, the ratio of reactance of the inductor to that of the capacitor is :

Explanation

To solve the problem, we need to analyze the reactance of both the inductor and the capacitor at the given frequencies.

### Step-by-Step Solution:

1. **Understand Reactance**:
- The reactance of an inductor (X_L) is given by the formula:
\[ X_L = \omega L \]
- The reactance of a capacitor (X_C) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]

2. **Given Condition at Frequency ω₁**:
- At frequency ω₁, the reactance of the inductor equals the reactance of the capacitor:
\[ X_L = X_C \]
- Substituting the formulas, we have:
\[ \omega_1 L = \frac{1}{\omega_1 C} \]

3. **Rearranging the Equation**:
- From the above equation, we can rearrange it to find a relationship between L and C:
\[ \omega_1^2 LC = 1 \]

4. **Change of Frequency to ω₂**:
- Now, we change the frequency to ω₂ = 2ω₁. We need to find the new reactances:
- For the inductor at ω₂:
\[ X_L(ω_2) = \omega_2 L = 2\omega_1 L \]
- For the capacitor at ω₂:
\[ X_C(ω_2) = \frac{1}{\omega_2 C} = \frac{1}{2\omega_1 C} \]

5. **Calculating the Ratio of Reactances**:
- Now, we calculate the ratio of the reactance of the inductor to that of the capacitor at ω₂:
\[ \frac{X_L(ω_2)}{X_C(ω_2)} = \frac{2\omega_1 L}{\frac{1}{2\omega_1 C}} = 2\omega_1 L \cdot 2\omega_1 C = 4\omega_1^2 LC \]

6. **Substituting the Relationship from Step 3**:
- From step 3, we know that ω₁² LC = 1. Thus:
\[ 4\omega_1^2 LC = 4 \cdot 1 = 4 \]

7. **Final Result**:
- Therefore, the ratio of the reactance of the inductor to that of the capacitor at frequency ω₂ is:
\[ \frac{X_L(ω_2)}{X_C(ω_2)} = 4 \]

### Conclusion:
The ratio of the reactance of the inductor to that of the capacitor when the frequency is changed to ω₂ = 2ω₁ is **4**.