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(B) The given equation of motion is $y = y_{0} \sin^{2} \omega t$.Using the trigonometric identity $\sin^{2} 	heta = \frac{1 - \cos 2	heta}{2}$,we c

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$\sqrt{\frac{g}{2y_{0}}}$

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(B) The given equation of motion is $y = y_{0} \sin^{2} \omega t$.Using the trigonometric identity $\sin^{2} 	heta = \frac{1 - \cos 2	heta}{2}$,we can rewrite the equation as:$y = \frac{y_{0}}{2} (1 - \cos 2\omega t)$$y - \frac{y_{0}}{2} = -\frac{y_{0}}{2} \cos 2\omega t$This represents simple harmonic motion about the equilibrium position $y_{eq} = \frac{y_{0}}{2}$ with an amplitude $A = \frac{y_{0}}{2}$.For a vertical spring-mass system,the equilibrium position is at a distance $y_{eq} = \frac{mg}{k}$ from the unstretched position.Therefore,$\frac{y_{0}}{2} = \frac{mg}{k}$,which implies $\frac{k}{m} = \frac{2g}{y_{0}}$.The angular frequency of the oscillation is given by $\Omega = \sqrt{\frac{k}{m}}$.From the equation $y = \frac{y_{0}}{2} - \frac{y_{0}}{2} \cos 2\omega t$,the angular frequency of the motion is $2\omega$.Thus,$2\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2g}{y_{0}}}$.Solving for $\omega$,we get $\omega = \frac{1}{2} \sqrt{\frac{2g}{y_{0}}} = \sqrt{\frac{2g}{4y_{0}}} = \sqrt{\frac{g}{2y_{0}}}$.

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 the unstretched spring. Then $\omega$ is

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