A spring mass system (mass $$m$$, spring constant $$k$$ and natural length $$l$$) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system rotates about it's axis with an angular velocity $$\omega$$, $$(k >> m\omega^2)$$ the relative change in the length of the spring is best given by the option:

Let the natural (unstretched) length of the spring be $$l$$. Because the free end is fixed at the centre of the disc, the mass of the spring-mass system always lies on a radius of the disc. When the disc starts rotating with a constant angular velocity $$\omega$$, the mass experiences an outward centrifugal force.

In the rotating (non-inertial) frame the only two radial forces on the mass are (1) the restoring force of the spring, directed towards the centre and equal in magnitude to $$k\Delta l$$, where $$\Delta l$$ is the extension, and (2) the centrifugal force, directed away from the centre and equal in magnitude to $$m\omega^{2}r$$, where $$r$$ is the distance of the mass from the centre after extension.

At the new static equilibrium these two forces must balance, so we write the condition $$k\Delta l = m\omega^{2}r.$$

The new length of the spring is clearly $$r = l + \Delta l.$$

Substituting this expression for $$r$$ into the force-balance equation we obtain $$k\Delta l = m\omega^{2}(l + \Delta l).$$

We now wish to express every quantity in terms of the natural length $$l$$. Define the relative (fractional) change in length as $$x = \frac{\Delta l}{l}.$$

Then $$\Delta l = xl,$$ and substituting this into the previous equation gives $$k(xl) = m\omega^{2}\bigl(l + xl\bigr).$$

We can cancel the common factor $$l$$ on both sides:

$$kx = m\omega^{2}(1 + x).$$

To isolate $$x$$, first divide both sides by $$k$$:

$$x = \frac{m\omega^{2}}{k}(1 + x).$$

Now expand the right-hand side and bring all terms involving $$x$$ to the left:

$$x - \frac{m\omega^{2}}{k}x = \frac{m\omega^{2}}{k}.$$

Factor out $$x$$ on the left:

$$x\left(1 - \frac{m\omega^{2}}{k}\right) = \frac{m\omega^{2}}{k}.$$

Finally, divide by the factor in parentheses:

$$x = \frac{\dfrac{m\omega^{2}}{k}}{1 - \dfrac{m\omega^{2}}{k}}.$$

The problem statement tells us that $$k \gg m\omega^{2}$$, which means $$\frac{m\omega^{2}}{k} \ll 1.$$ For such a small quantity we can safely neglect its square and higher powers, so we approximate $$\frac{1}{1 - \dfrac{m\omega^{2}}{k}} \approx 1.$$

Under this first-order approximation the relative change reduces to

$$x \approx \frac{m\omega^{2}}{k}.$$

Because $$x$$ was defined as $$\dfrac{\Delta l}{l}$$, we have found

$$\frac{\Delta l}{l} \approx \frac{m\omega^{2}}{k}.$$

This expression exactly matches Option C.

Hence, the correct answer is Option 3.