Answer:
Displacement, \[y={{\sin }^{2}}\omega t\] Velocity, \[\upsilon =\frac{dy}{dt}2\sin \omega t\times \cos \omega t\times \omega \] \[=\omega \sin 2\omega t\] Acceleration, \[a=\frac{d\upsilon }{dt}=\omega \times \cos 2\omega t\times 2\omega \] \[=2{{\omega }^{2}}\cos 2\omega t\] As the acceleration a is not proportional to displacement y, the given function does not represent SHM. It represents a periodic motion of angular frequency \[2\omega \]. \[\therefore \] Time period, \[T=\frac{2\pi }{\text{Angular frequency}}=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }.\]
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