A particle moves along X-axis according to the law $x=A{\sin }^2(\omega t-\frac{\pi }{4})$. Its velocity is v. Choose the correct option(s).

$$\begin{gathered} A=A{\sin }^2(\omega t-\frac{\pi }{4}) \\ =A\left[\frac{1-\cos 2(\omega t-\frac{\pi }{4})}{2}\right]=\frac{A}{2}-\frac{A}{2}\cos (2\omega t-\frac{\pi }{2}) \\ or x=\frac{A}{2}-\frac{A}{2}\sin 2\omega t\_\_\_\_\_\left(i\right) \end{gathered}$$

$\therefore$Amplitude $=\frac{A}{2}$, angular frequency $=2\omega$

$$\begin{gathered} v=\frac{dx}{dt}=-\frac{A}{2}2\omega \cos 2\omega t \\ or v=-A\omega \cos 2\omega t \end{gathered}$$

From Eq.(i), $\frac{A}{2}\sin 2\omega t=\frac{A}{2}-x$

or  $\sin 2\omega t=1-\frac{2}{A}x$

$$\begin{gathered} \therefore \cos 2\omega t=\sqrt{1-{(1-\frac{2}{A}x)}^2} \\ so, v=-A\omega \sqrt{1-{(1-\frac{2}{A}x)}^2} \\ {v}^2=A^2{\omega }^2[1-{(1-\frac{2}{A}x)}^2] \\ \therefore v^2=4{\omega }^{2}x(A-x) \end{gathered}$$