A particle of mass `m` is fixed to one end of a massless spring of spring constant `k` and natural length `l_(0)`. The system is rotated about the other end of the spring with an angular velocity `omega` ub gravity-free space. The final length of spring is

Options

A. `m omega^(2) l)/(k)`
B. `(m omega^(2) l)/(k-m omega^(2))`
C. `(m omega^(2) l)/(k + m omega^(2))`
D. none of these

Explanation

To find the final length of the spring when a particle of mass `m` is attached to one end of a massless spring of spring constant `k` and natural length `l_0`, and the system is rotated with an angular velocity `ω` in gravity-free space, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Mass**:
- When the mass is rotated, it experiences a centrifugal force acting outward due to the rotation. This force can be expressed as:
\[ F_{\text{centrifugal}} = m \omega^2 r \]
- Here, `r` is the distance from the fixed end of the spring to the mass, which is the final length of the spring (`l_f`).

2. **Define the Final Length of the Spring**:
- The final length of the spring can be expressed as:
\[ r = l_0 + x \]
- Where `x` is the elongation (stretch) of the spring from its natural length `l_0`.

3. **Spring Force**:
- The spring exerts a restoring force given by Hooke's Law:
\[ F_{\text{spring}} = kx \]

4. **Set Up the Equilibrium Condition**:
- In equilibrium, the centrifugal force is balanced by the spring force:
\[ kx = m \omega^2 r \]

5. **Substituting for `r`**:
- Substitute `r` in the equilibrium equation:
\[ kx = m \omega^2 (l_0 + x) \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ kx - m \omega^2 x = m \omega^2 l_0 \]
- Factor out `x`:
\[ x(k - m \omega^2) = m \omega^2 l_0 \]

7. **Solve for `x`**:
- Solving for `x` gives:
\[ x = \frac{m \omega^2 l_0}{k - m \omega^2} \]

8. **Calculate the Final Length of the Spring**:
- The final length of the spring `l_f` is:
\[ l_f = l_0 + x \]
- Substitute the expression for `x`:
\[ l_f = l_0 + \frac{m \omega^2 l_0}{k - m \omega^2} \]

9. **Simplifying the Final Length**:
- Factor out `l_0`:
\[ l_f = l_0 \left(1 + \frac{m \omega^2}{k - m \omega^2}\right) \]
- Combine the terms:
\[ l_f = l_0 \left(\frac{k}{k - m \omega^2}\right) \]

### Final Result:
Thus, the final length of the spring is:
\[ l_f = \frac{k l_0}{k - m \omega^2} \]