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(D) The given equation is $x = x_0 \sin^2 \omega t$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we can rewrite the e

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With time period $(\pi / \omega)$

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(D) The given equation is $x = x_0 \sin^2 \omega t$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we can rewrite the equation as:$x = x_0 \left( \frac{1 - \cos 2\omega t}{2} ight) = \frac{x_0}{2} - \frac{x_0}{2} \cos 2\omega t$.This represents a simple harmonic motion shifted by a constant displacement $\frac{x_0}{2}$.The standard form of $SHM$ is $x' = A \cos(\omega' t + \phi)$,where $x' = x - \frac{x_0}{2}$.Here,the amplitude $A = \frac{x_0}{2}$ and the angular frequency $\omega' = 2\omega$.The time period $T$ is given by $T = \frac{2\pi}{\omega'} = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.

$A$ particle moves on the $x-$axis according to the equation $x = x_0 \sin^2 \omega t$. Is the motion simple harmonic?

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