When a particle of mass $$m$$ is attached to a vertical spring of spring constant $$k$$ and released, its motion is described by $$y(t) = y_0\sin^2\omega t$$, where 'y' is measured from the lower end of unstretched spring. Then $$\omega$$ is:

Let the downward direction be taken as positive. The mass is attached to the lower end of the spring, and the coordinate $$y$$ is measured downward from the unstretched (natural) length of the spring. Hence the restoring force of the spring is upward and has magnitude $$ky$$, while the weight $$mg$$ acts downward.

From Newton’s second law we therefore have

$$m\,\dfrac{d^{2}y}{dt^{2}} = mg - ky.$$

Dividing by $$m$$ gives the linear differential equation

$$\dfrac{d^{2}y}{dt^{2}} + \dfrac{k}{m}\,y = g. \quad -(1)$$

The problem states that the displacement as a function of time is

$$y(t)=y_{0}\sin^{2}\!\bigl(\omega t\bigr).$$

We shall substitute this trial solution into equation (1) and determine the value of $$\omega$$ that makes the equation an identity for all times.

First we rewrite the given function with a double-angle expression:

$$y(t)=y_{0}\sin^{2}\!\bigl(\omega t\bigr) =y_{0}\,\dfrac{1-\cos\!\bigl(2\omega t\bigr)}{2} =\dfrac{y_{0}}{2}-\dfrac{y_{0}}{2}\cos\!\bigl(2\omega t\bigr). \quad -(2)$$

Now we differentiate. From the chain rule, the first derivative is

$$\dfrac{dy}{dt}=y_{0}\,2\sin\!\bigl(\omega t\bigr)\cos\!\bigl(\omega t\bigr)\,\omega =y_{0}\omega\sin\!\bigl(2\omega t\bigr).$$

Differentiating once more, the second derivative becomes

$$\dfrac{d^{2}y}{dt^{2}} = y_{0}\omega\, 2\omega\cos\!\bigl(2\omega t\bigr) =2y_{0}\omega^{2}\cos\!\bigl(2\omega t\bigr). \quad -(3)$$

We now substitute equations (2) and (3) into the equation of motion (1).

$$2y_{0}\omega^{2}\cos\!\bigl(2\omega t\bigr) +\dfrac{k}{m}\left[\dfrac{y_{0}}{2}-\dfrac{y_{0}}{2}\cos\!\bigl(2\omega t\bigr)\right]=g.$$

Expanding and collecting like terms yields

$$2y_{0}\omega^{2}\cos\!\bigl(2\omega t\bigr) +\dfrac{ky_{0}}{2m} -\dfrac{ky_{0}}{2m}\cos\!\bigl(2\omega t\bigr)=g.$$

The above identity must hold for every value of $$t$$. Therefore the coefficient of the time-dependent term $$\cos\!\bigl(2\omega t\bigr)$$ must vanish, and the remaining constant part must equal $$g$$.

Setting the constant part equal to $$g$$ gives

$$\dfrac{ky_{0}}{2m}=g \;\;\Longrightarrow\;\; \dfrac{k}{m}= \dfrac{2g}{y_{0}}. \quad -(4)$$

Setting the coefficient of $$\cos\!\bigl(2\omega t\bigr)$$ to zero gives

$$2y_{0}\omega^{2}-\dfrac{ky_{0}}{2m}=0.$$

Dividing by $$y_{0}$$ and substituting $$\dfrac{k}{m}$$ from equation (4), we get

$$2\omega^{2}-\dfrac{1}{2}\left(\dfrac{2g}{y_{0}}\right)=0 \;\;\Longrightarrow\;\; 2\omega^{2}-\dfrac{g}{y_{0}}=0.$$

Solving for $$\omega^{2}$$, we find

$$\omega^{2}=\dfrac{g}{2y_{0}} \;\;\Longrightarrow\;\; \omega=\sqrt{\dfrac{g}{2y_{0}}}.$$

This coincides with Option C in the given list.

Hence, the correct answer is Option C.