Match the List I with List II:

Choose the correct answer from the options given below: 

A. $$\sin^2\omega t$$ can be rewritten as $$\frac{1}{2} - \frac{1}{2}\cos(2\omega t)$$. Since it consists of a single sinusoidal term, it represents SHM with a time period $$T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$$ $$\rightarrow \text{III}$$

B. $$\sin^3(2\omega t)$$ expands to $$\frac{3}{4}\sin(2\omega t) - \frac{1}{4}\sin(6\omega t)$$. The combination of multiple distinct harmonic frequencies means it is not SHM. Its fundamental period is governed by the lowest frequency ($$2\omega$$), giving $$T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$$ $$\rightarrow \text{I}$$

C. $$\sin(\omega t) + \cos(\pi\omega t)$$, the individual periods are $$T_1 = \frac{2\pi}{\omega}$$ and $$T_2 = \frac{2}{\omega}$$. Their ratio $$\frac{T_1}{T_2} = \pi$$ is an irrational number, which makes the combined function non-periodic $$\rightarrow \text{IV}$$

D. $$\cos\omega t + \cos 2\omega t$$ contains two separate harmonic frequencies ($$\omega$$ and $$2\omega$$), so it is not SHM. The overall time period is the LCM of their individual periods ($$\frac{2\pi}{\omega}$$ and $$\frac{\pi}{\omega}$$), which equals $$\frac{2\pi}{\omega}$$ $$\rightarrow \text{II}$$