A mass m attached to spring of natural length `l_0` and spring constant k. One end of string is attached to centre of disc in horizontal plane which is being rotated by constant angular speed `omega`. Find extension per unit length in spring (given `k gtgt m omega^2` ) :

Options

A. `(momega^(2))/(2k)`
B. `(2momega^(2))/(k)`
C. `(momega^(2))/(k)`
D. `sqrt((2)/(3))((momega^(2))/(k))`

Answer

To solve the problem, we need to find the extension per unit length in the spring when a mass \( m \) is attached to it and the spring is subjected to a centrifugal force due to the rotation of a disc at a constant angular speed \( \omega \).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Mass:**
- The mass \( m \) experiences a centrifugal force due to the rotation of the disc, which can be expressed as:
\[ F_{\text{centrifugal}} = m \omega^2 r \]
- Here, \( r \) is the total length from the center of the disc to the mass, which is given by:
\[ r = l_0 + x \]
- Where \( l_0 \) is the natural length of the spring and \( x \) is the extension of the spring.

2. **Write the Equation for the Spring Force:**
- The spring exerts a restoring force given by Hooke's Law:
\[ F_{\text{spring}} = kx \]
- Where \( k \) is the spring constant.

3. **Set Up the Equation of Motion:**
- At equilibrium, the centrifugal force is balanced by the spring force:
\[ kx = m \omega^2 (l_0 + x) \]

4. **Rearranging the Equation:**
- Expanding the right side gives:
\[ kx = m \omega^2 l_0 + m \omega^2 x \]
- Rearranging this equation leads to:
\[ kx - m \omega^2 x = m \omega^2 l_0 \]
- Factoring out \( x \) from the left side:
\[ x(k - m \omega^2) = m \omega^2 l_0 \]

5. **Solve for \( x \):**
- We can solve for \( x \):
\[ x = \frac{m \omega^2 l_0}{k - m \omega^2} \]

6. **Find the Extension per Unit Length:**
- We need to find the extension per unit length, which is \( \frac{x}{l_0} \):
\[ \frac{x}{l_0} = \frac{m \omega^2}{k - m \omega^2} \]

7. **Apply the Given Condition \( k \gg m \omega^2 \):**
- Since \( k \) is much greater than \( m \omega^2 \), we can approximate:
\[ k - m \omega^2 \approx k \]
- Therefore, the expression simplifies to:
\[ \frac{x}{l_0} \approx \frac{m \omega^2}{k} \]

### Final Result:
The extension per unit length in the spring is given by:
\[ \frac{x}{l_0} = \frac{m \omega^2}{k} \]