The displacement of a particle is given at time $t$, by:$x=A \sin (-2 \omega t)+B \sin ^2 \omega t \quad$ Then,
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(a)
The displacement of the particle is given by:
$x =A \sin (-2 \omega t)+B \sin ^2 \omega t$
$=-A \sin 2 \omega t+\frac{B}{2}(1-\cos 2 \omega t)$
$=-\left(A \sin 2 \omega t+\frac{B}{2} \cos 2 \omega t\right)+\frac{B}{2}$
This motion represents SHM with an amplitude: $\sqrt{A^2+\frac{B^2}{4}}$, and mean position $\frac{B}{2}$
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