Two identical particles each of mass m are connected by a light spring of stiffness K. The initial deformation of the spring when the system was at rest, `x_(0)`. After releasing the system, if one of the particles has a speed v and the deformation of the spring is x, find the speed of the other particle neglecting friction.

Answer

To solve the problem, we will use the principle of conservation of mechanical energy. The initial potential energy stored in the spring when it is deformed by \( x_0 \) will be equal to the sum of the kinetic energies of the two particles and the potential energy stored in the spring when it is deformed by \( x \).

### Step-by-Step Solution:

1. **Identify Initial and Final States**:
- Initially, the spring is deformed by \( x_0 \) and both particles are at rest. Thus, the initial kinetic energy is 0.
- After release, one particle has a speed \( v \) and the spring is deformed by \( x \).

2. **Calculate Initial Energy**:
- The initial potential energy \( U_i \) stored in the spring when it is deformed by \( x_0 \) is given by:
\[ U_i = \frac{1}{2} K x_0^2 \]

3. **Calculate Final Energy**:
- The final energy \( U_f \) consists of the kinetic energy of both particles and the potential energy stored in the spring when it is deformed by \( x \):
\[ U_f = \frac{1}{2} m v^2 + \frac{1}{2} m v_1^2 + \frac{1}{2} K x^2 \]
- Here, \( v_1 \) is the speed of the other particle which we need to find.

4. **Apply Conservation of Energy**:
- According to the conservation of mechanical energy:
\[ U_i = U_f \]
- Substituting the expressions for initial and final energy:
\[ \frac{1}{2} K x_0^2 = \frac{1}{2} m v^2 + \frac{1}{2} m v_1^2 + \frac{1}{2} K x^2 \]

5. **Simplify the Equation**:
- Multiply through by 2 to eliminate the fractions:
\[ K x_0^2 = m v^2 + m v_1^2 + K x^2 \]
- Rearranging gives:
\[ K x_0^2 - K x^2 = m v^2 + m v_1^2 \]

6. **Solve for \( v_1 \)**:
- Rearranging the equation to isolate \( v_1^2 \):
\[ m v_1^2 = K x_0^2 - K x^2 - m v^2 \]
- Dividing by \( m \):
\[ v_1^2 = \frac{K}{m} (x_0^2 - x^2) - v^2 \]
- Taking the square root to find \( v_1 \):
\[ v_1 = \sqrt{\frac{K}{m} (x_0^2 - x^2) - v^2} \]

### Final Answer:
The speed of the other particle is:
\[ v_1 = \sqrt{\frac{K}{m} (x_0^2 - x^2) - v^2} \]