The function of time representing a simple harmonic motion with a period of \(\frac{\pi}{\omega}\) is :

\(\frac{\pi}{\omega}=\frac{2 \pi}{\omega^{\prime}}\)

\(\omega^{\prime}=2 \omega \rightarrow\) 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.