Physics - SHM Question with Solution | TestHub

$\text{y}={\text{sin}}^2\omega t=\frac{1-\text{cos}2\omega t}{2}=\frac{1}{2}-\frac{\text{cos}2\omega t}{2 }$


It is a periodic motion but it is not SHM


$\therefore$Angular speed =$2\omega$

$\therefore$Period$\mathit{\text{T}}=\frac{2\pi }{\text{angular speed}}=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$


Hence option ( a periodic, but not simple harmonic motion with a period$\pi /\omega$)  represents the answer.