The function of time representing a simple harmonic motion with a period of $\frac{\pi}{\omega}$ is :
JEE MAIN 2021, Diffcult
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Time period $T =\frac{2 \pi}{\omega^{\prime}}$

$\frac{\pi}{\omega}=\frac{2 \pi}{\omega^{\prime}}$

$\omega^{\prime}=2 \omega \rightarrow$ Angular frequency of $SHM$

Option $(3):$

$\sin ^{2} \omega t=\frac{1}{2}\left(2 \sin ^{2} \omega t\right)=\frac{1}{2}(1-\cos 2 \omega t)$

Angular frequency of $\left(\frac{1}{2}-\frac{1}{2} \cos 2 \omega t\right)$ is $2 \omega$

Option $(4):$

Angular frequency of $SHM$

$3 \cos \left(\frac{\pi}{4}-2 \omega t\right)$ is $2 \omega$

So option $(3)\, \& (4)$ both have angular frequency $2 \omega$ but option $(4)$ is direct answer.

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