One end of a massless spring of spring constant $$k$$ and natural length $$l_0$$ is fixed while the other end is connected to a small object of mass $$m$$ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $$\omega$$ about an axis passing through fixed end, then the elongation of the spring will be

A massless spring of spring constant $$k$$ and natural length $$l_0$$ has one end fixed. A mass $$m$$ at the other end rotates on a frictionless table with angular velocity $$\omega$$.

Set up the equation of motion.

Let the elongation of the spring be $$x$$. The total length of the spring is $$l_0 + x$$.

The mass moves in a circle of radius $$r = l_0 + x$$.

The centripetal force required is provided by the spring force:

$$kx = m\omega^2(l_0 + x)$$

Solve for $$x$$.

$$kx = m\omega^2 l_0 + m\omega^2 x$$

$$kx - m\omega^2 x = m\omega^2 l_0$$

$$x(k - m\omega^2) = m\omega^2 l_0$$

$$x = \frac{m\omega^2 l_0}{k - m\omega^2}$$

The elongation of the spring is $$\dfrac{m\omega^2 l_0}{k - m\omega^2}$$.

The correct answer is Option C.