One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacment, the work done by the spring is `+(1/2)kx^(2)`. The possible cases are.

Options

A. The spring was initially compessed by a distance x and was finally in its natural length .
B. It was initially stretched by a distance x and finally was in its natural length.
C. It was initially natural length and finally in a compressed position.
D. It was initially in its natural length and finally in a stretched position.

Explanation

To solve the problem, we need to analyze the work done by a spring in different scenarios involving its compression and elongation. The work done by the spring is given by the formula:

\[ W = \frac{1}{2} k x^2 \]

where \( W \) is the work done, \( k \) is the spring constant, and \( x \) is the displacement from the spring's natural length.

### Step-by-Step Solution:

1. **Understand the Spring's Behavior**:
The spring can either be compressed or stretched. The work done by the spring depends on the initial and final positions of the spring.

2. **Analyze Each Case**:
- **Case 1**: The spring is initially compressed by a distance \( x \) and then returns to its natural length.
- Here, the spring does positive work as it moves from a compressed state to its natural length.
- Work done: \( W = \frac{1}{2} k x^2 \) (positive).

- **Case 2**: The spring is initially stretched by a distance \( x \) and then returns to its natural length.
- Similar to Case 1, the spring does positive work as it moves from a stretched state to its natural length.
- Work done: \( W = \frac{1}{2} k x^2 \) (positive).

- **Case 3**: The spring is initially at its natural length and then is compressed.
- In this case, the spring is being compressed, and it stores potential energy. However, the work done by the spring is not applicable here since it is not returning to a natural position.
- Work done: \( W = 0 \) (no displacement from natural length).

- **Case 4**: The spring is initially at its natural length and then is stretched.
- Similar to Case 3, the spring is being stretched, and it stores potential energy but does not perform work on the block as it is not returning to a natural position.
- Work done: \( W = 0 \) (no displacement from natural length).

3. **Conclusion**:
The only cases where the spring does work on the block are Case 1 and Case 2. Therefore, the possible cases where work is done by the spring are:
- Case 1: Initially compressed to natural length.
- Case 2: Initially stretched to natural length.

### Final Answer:
The correct options are **Case 1 and Case 2**.

---