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

Time period $T=\frac{2\pi }{\omega \text{'}}$

$\frac{\pi }{\omega }=\frac{2\pi }{\omega \text{'}}$

$\omega \text{'}=2\omega \to$ Angular frequency of SHM

Option $\left(3\right)$

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

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

Option $\left(4\right)$

Angular frequency of SHM

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

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