1. Set up the situation

A mass $m$ is attached to a light spring of:

• force constant $k$
• natural length $\ell$

The other end is fixed, and the mass rotates in a circle with angular speed $\omega$ in gravity-free space.

Let the extension in the spring be $x$.

Then the actual length of the spring, which is also the radius of circular motion, is $r=\ell+x.$

2. Forces acting on the particle

Since there is no gravity, the only force on the particle is the spring force toward the fixed end.

Magnitude of spring force: $F_s = kx.$

For circular motion of radius $r=\ell+x$ and angular speed $\omega$, the required centripetal force is $F_c = m\omega^2 r = m\omega^2(\ell+x).$

Since the spring provides the centripetal force, $kx = m\omega^2(\ell+x).$

3. Solve for the extension $x$

Expand the right-hand side: $kx = m\omega^2\ell + m\omega^2x.$

Bring terms involving $x$ to one side: $kx - m\omega^2x = m\omega^2\ell.$

Factor out $x$: $x(k-m\omega^2)=m\omega^2\ell.$

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

4. Match with the options

This matches: $\boxed{\text{A: } \frac{m\ell\omega^2}{k-m\omega^2}}$

5. Comparison with stored correct answer

Stored correct answer: A

Our derived answer is also A, so they agree.