A particle of mass 'm' attached with one end of a light spring of stiffness (k) is pushed with a speed (v) on a smooth horizontal plane. If `l_(0)=` natural length of the spring and `x_(0)` = elongation of the spring at the given instant.

Options

A. `F_(c.p)= kx_(0)`
B. `F_(c.p)= (mv^(2))/(l_(0)+x_(0))`
C. `F_(c.p) = kx_(0)- (mv^(2))/(l_(0)+x_(0))`
D. All of the above

Answer

To solve the problem step by step, we will analyze the forces acting on the particle attached to the spring and derive the necessary equations.

### Step-by-Step Solution:

1. **Identify the Forces:**
The particle of mass 'm' is moving with a speed 'v' on a smooth horizontal plane. The forces acting on the particle are the centripetal force required for circular motion and the restoring force exerted by the spring.

2. **Centripetal Force:**
The centripetal force \( F_c \) required to keep the particle moving in a circular path is given by the formula:
\[ F_c = \frac{mv^2}{r} \]
where \( r \) is the radius of the circular path.

3. **Determine the Radius:**
The radius \( r \) of the circular motion can be expressed in terms of the natural length of the spring \( l_0 \) and the elongation \( x_0 \) of the spring:
\[ r = l_0 + x_0 \]

4. **Restoring Force of the Spring:**
The restoring force \( F_r \) exerted by the spring when it is elongated by \( x_0 \) is given by Hooke's Law:
\[ F_r = kx_0 \]
where \( k \) is the stiffness of the spring.

5. **Set the Forces Equal:**
For the particle to maintain circular motion, the centripetal force must equal the restoring force of the spring:
\[ \frac{mv^2}{r} = kx_0 \]
Substituting \( r = l_0 + x_0 \) into the equation gives:
\[ \frac{mv^2}{l_0 + x_0} = kx_0 \]

6. **Final Equation:**
Rearranging the equation allows us to express the relationship between the mass, speed, spring constant, and elongation:
\[ mv^2 = kx_0(l_0 + x_0) \]

### Conclusion:
The equation \( mv^2 = kx_0(l_0 + x_0) \) relates the mass of the particle, its speed, the spring constant, and the elongation of the spring.