Two springs have spring constants `k_(1)and k_(2) (k_(1)nek_(2)).` Both are extended by same force. If their elastic potential energical are `U_(1)and U_(2),` then `U_(1)`:`U_(2)` is

Options

A. `(K_(1))/(K_(2))`
B. `(K_(2))/(K_(1))`
C. `sqrt((K_(1))/(K_(2)))`
D. `sqrt((K_(2))/(K_(2)))`

Answer

To solve the problem, we need to find the ratio of the elastic potential energies \( U_1 \) and \( U_2 \) of two springs with spring constants \( k_1 \) and \( k_2 \) when they are extended by the same force \( F \).

### Step-by-Step Solution:

1. **Understanding Elastic Potential Energy**:
The elastic potential energy \( U \) stored in a spring is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the extension of the spring.

2. **Relating Force and Extension**:
The force \( F \) applied to the spring is related to the spring constant and the extension by Hooke's Law:
\[ F = k x \]
Rearranging this gives:
\[ x = \frac{F}{k} \]

3. **Substituting for Extension**:
Now, we can substitute \( x \) back into the potential energy formula. For spring 1 with spring constant \( k_1 \):
\[ U_1 = \frac{1}{2} k_1 \left(\frac{F}{k_1}\right)^2 = \frac{1}{2} k_1 \cdot \frac{F^2}{k_1^2} = \frac{F^2}{2 k_1} \]
For spring 2 with spring constant \( k_2 \):
\[ U_2 = \frac{1}{2} k_2 \left(\frac{F}{k_2}\right)^2 = \frac{1}{2} k_2 \cdot \frac{F^2}{k_2^2} = \frac{F^2}{2 k_2} \]

4. **Finding the Ratio of Potential Energies**:
Now, we can find the ratio of the potential energies \( U_1 \) and \( U_2 \):
\[ \frac{U_1}{U_2} = \frac{\frac{F^2}{2 k_1}}{\frac{F^2}{2 k_2}} = \frac{2 k_2}{2 k_1} = \frac{k_2}{k_1} \]

5. **Conclusion**:
Therefore, the ratio of the elastic potential energies \( U_1 : U_2 \) is:
\[ U_1 : U_2 = k_2 : k_1 \]

### Final Answer:
The ratio \( U_1 : U_2 = k_2 : k_1 \).