A spring of force constant `1200 Nm^(-1)` is mounted on a horizontal table and a mass of 3.0 kg of attached to the free and of the spring. The mass is made to rotate with an angular velocity of 10 rad/s. What is the elongation produced in the spring if its unstretched length is 60 cm ?

Explanation

To solve the problem step by step, we will follow the physics principles related to circular motion and Hooke's law for springs.

### Given Data:
- Force constant of the spring, \( k = 1200 \, \text{N/m} \)
- Mass attached to the spring, \( m = 3.0 \, \text{kg} \)
- Angular velocity, \( \omega = 10 \, \text{rad/s} \)
- Unstretched length of the spring, \( L_0 = 60 \, \text{cm} = 0.6 \, \text{m} \)

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