In case of an ideal spring, we take the reference point (where potential energy is assumed to be zero) at extension of `x_(0)` instead of taking at natural length. Then potential energy when spring is extended by `x` is `:`

Options

A. `(1)/(2)k(x-x_(0))^(2)`
B. `(1)/(2)k(x^(2)-x_(0)^(2))`
C. `(1)/(2)k(x^(2)-2x x_(0))`
D. `(1)/(2)k(2x x_(0)-x^(2))`

Explanation

To solve the problem, we need to determine the potential energy of an ideal spring when it is extended by a distance \( x \) from a reference point where the potential energy is considered to be zero, which is at an extension of \( x_0 \).

### Step-by-Step Solution:

1. **Understanding the Spring Potential Energy Formula**:
The potential energy \( PE \) stored in a spring when it is extended or compressed by a distance \( x \) from its natural length is given by the formula:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant.

2. **Setting the Reference Point**:
In this case, the reference point where the potential energy is zero is at an extension of \( x_0 \). This means that when the spring is extended by \( x_0 \), the potential energy is defined as zero.

3. **Calculating Initial Potential Energy**:
When the spring is extended by \( x_0 \), the potential energy at this point is:
\[ PE_{initial} = \frac{1}{2} k x_0^2 \]
However, since we are taking this as our reference point, we consider this energy to be zero.

4. **Calculating Final Potential Energy**:
Now, when the spring is extended by \( x \), the potential energy at this extension is:
\[ PE_{final} = \frac{1}{2} k x^2 \]

5. **Finding the Change in Potential Energy**:
The change in potential energy when the spring is extended from \( x_0 \) to \( x \) is given by:
\[ \Delta PE = PE_{final} - PE_{initial} \]
Since \( PE_{initial} = 0 \) (as defined), we have:
\[ \Delta PE = \frac{1}{2} k x^2 - \frac{1}{2} k x_0^2 \]

6. **Final Expression**:
Therefore, the potential energy when the spring is extended by \( x \) from the reference point \( x_0 \) is:
\[ PE = \frac{1}{2} k (x^2 - x_0^2) \]

### Final Answer:
The potential energy when the spring is extended by \( x \) is:
\[ PE = \frac{1}{2} k (x^2 - x_0^2) \]