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

A simple harmonic motion with period $$\frac{\pi}{\omega}$$ corresponds to an angular frequency of $$\frac{2\pi}{T} = \frac{2\pi}{\pi/\omega} = 2\omega$$.

Option (1): $$\sin(\omega t) + \cos(\omega t) = \sqrt{2}\sin(\omega t + \pi/4)$$, which has angular frequency $$\omega$$ and period $$\frac{2\pi}{\omega}$$. This does not match.

Option (2): $$\cos(\omega t) + \cos(2\omega t) + \cos(3\omega t)$$ is a sum of terms with different frequencies, so it is not SHM at all.

Option (3): $$\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2}$$. Although it has angular frequency $$2\omega$$, the constant offset means it does not oscillate about the mean position in the standard SHM sense.

Option (4): $$3\cos\left(\frac{\pi}{4} - 2\omega t\right) = 3\cos\left(2\omega t - \frac{\pi}{4}\right)$$. This is a single cosine function with angular frequency $$2\omega$$, giving a period of $$\frac{2\pi}{2\omega} = \frac{\pi}{\omega}$$. This is a simple harmonic motion with the required period.