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Step-by-Step Solution
Step 1: Write down the general form of the equation of motion
For a mass attached to a vertical spring, the net force at a displacement
$y$ (from the unstretched position) is given by
$F = -k\,y + mg,$
which leads to the equation of motion
$m\,\ddot{y}(t) = -k\,y(t) + m\,g.$
Step 2: Express $y(t)$ and compute its second derivative
According to the problem, the motion is described by
$y(t) = y_{0}\,\sin^{2}\bigl(\omega t\bigr).$
Differentiate twice with respect to $t$:
First derivative (velocity):
$$
\dot{y}(t)
= \frac{d}{dt} \Bigl[y_{0}\,\sin^{2}(\omega t)\Bigr]
= y_{0}\,(2\,\sin(\omega t)\cos(\omega t))\,\omega
= y_{0}\,\omega\,\sin(2\,\omega t).
$$
Second derivative (acceleration):
$$
\ddot{y}(t)
= \frac{d}{dt}\Bigl[y_{0}\,\omega\,\sin(2\,\omega t)\Bigr]
= y_{0}\,\omega\,(2\,\omega\,\cos(2\,\omega t))
= 2\,y_{0}\,\omega^{2}\,\cos(2\,\omega t).
$$
Step 3: Apply initial conditions at $t = 0$
At $t = 0$, the mass starts from $y(0) = 0$ (the unstretched position).
Substitute $t=0$ into the differential equation
$$
\ddot{y}(0) = -\frac{k}{m}\,y(0) + g.
$$
We have:
$y(0) = y_{0}\,\sin^{2}(0) = 0.$
From the expression for $\ddot{y}(t)$,
$$
\ddot{y}(0) = 2\,y_{0}\,\omega^{2}\,\cos(0)
= 2\,y_{0}\,\omega^{2}.
$$
Thus, the left-hand side at $t=0$ is $2\,y_{0}\,\omega^{2}.$
On the right-hand side, $y(0) = 0$ implies $-\frac{k}{m}\cdot 0 + g = g.$
Therefore,
$$
2\,y_{0}\,\omega^{2} = g
\quad\Longrightarrow\quad
\omega^{2} = \frac{g}{2\,y_{0}}.
$$
Step 4: Write the final expression for $\omega$
Taking the positive square root for angular frequency,
$$
\omega = \sqrt{\frac{g}{2\,y_{0}}}.
$$
This matches the correct answer given.
Answer Verification
Hence, the acceptable expression for $\omega$ is
$$
\omega = \sqrt{\frac{g}{2\,y_{0}}}.
$$
