1. Interpret the given motion

The particle is attached to a vertical spring and its position is given by

$y(t)=y_0\sin^2(\omega t)$

Using the identity

$\sin^2(\omega t)=\frac{1-\cos(2\omega t)}{2}$

we can rewrite:

$y(t)=\frac{y_0}{2}-\frac{y_0}{2}\cos(2\omega t)$

This is of the form

$y(t)=\text{mean position} + \text{amplitude} \cdot \cos(2\omega t+\phi)$

So:

• mean position from the lower end of unstretched spring = $\dfrac{y_0}{2}$
• amplitude = $\dfrac{y_0}{2}$
• angular frequency of oscillation = $2\omega$

2. Use equilibrium condition for a vertical spring

For a mass $m$ hanging from a vertical spring, the equilibrium extension is

$x_{eq}=\frac{mg}{k}$

Since $y$ is measured from the lower end of the unstretched spring, the equilibrium position is exactly this extension.

From the given motion, the mean position is the equilibrium position, hence

$\frac{mg}{k}=\frac{y_0}{2}$

So,

$\frac{k}{m}=\frac{2g}{y_0}$

3. Relate this to angular frequency

For vertical spring-mass SHM, the actual angular frequency is

$\Omega=\sqrt{\frac{k}{m}}$

But from the expression $y(t)=y_0\sin^2(\omega t)$, we found that the oscillation occurs with angular frequency

$\Omega=2\omega$

Therefore,

$2\omega=\sqrt{\frac{k}{m}}$

Using $\dfrac{k}{m}=\dfrac{2g}{y_0}$,

$2\omega=\sqrt{\frac{2g}{y_0}}$

Hence,

$\omega=\frac{1}{2}\sqrt{\frac{2g}{y_0}}=\sqrt{\frac{g}{2y_0}}$

4. Match with options

$\omega=\sqrt{\frac{g}{2y_0}}$

So the correct option is:

D

5. Comparison with stored answer

Stored correct answer: D

This matches our derived answer.