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(B) Given,the instantaneous displacement of the particle is $x = A \sin^2(\omega t - \frac{\pi}{4})$.Using the trigonometric identity $\sin^2 	heta =

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$\frac{\pi}{\omega}$

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(B) Given,the instantaneous displacement of the particle is $x = A \sin^2(\omega t - \frac{\pi}{4})$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos(2	heta)}{2}$,we can rewrite the expression as:$x = A \left[ \frac{1 - \cos(2(\omega t - \frac{\pi}{4}))}{2} ight]$$x = \frac{A}{2} [1 - \cos(2\omega t - \frac{\pi}{2})]$In simple harmonic motion,the general form is $x = x_0 + A' \cos(\omega' t + \phi)$.Comparing the expression,the angular frequency of the oscillation is $\omega' = 2\omega$.The time period $T'$ is given by $T' = \frac{2\pi}{\omega'}$.Substituting $\omega' = 2\omega$,we get $T' = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.

$A$ particle is executing simple harmonic motion with an instantaneous displacement $x = A \sin^2(\omega t - \frac{\pi}{4})$. The time period of oscillation of the particle is

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