Match List - I with List - II.
$$\begin{array}{|l|l|l|l|}\hline & \text{List - I} & & \text{List - II} \\ \hline \text{A.} & \sin^2\omega t & \text{I.} & \text{Periodic with time period } T = \frac{\pi}{\omega} \text{ but not simple harmonic motion (SHM)} \\ \hline \text{B.} & \sin^3(2\omega t) & \text{II.} & \text{Periodic with time period } T = \frac{2\pi}{\omega} \text{ but Not SHM} \\ \hline \text{C.} & \sin(\omega t) + \cos(\pi\omega t) & \text{III.} & \text{Periodic with time period } T = \frac{\pi}{\omega} \text{ and SHM} \\ \hline \text{D.} & \cos\omega t + \cos 2\omega t & \text{IV.} & \text{Non-periodic} \\ \hline \end{array}$$
Choose the correct answer from the options given below :
Answer: (A) A-III, B-I, C-IV, D-II
- A: $\sin^2\omega t = \dfrac{1 - \cos 2\omega t}{2}$, which is SHM about $\frac{1}{2}$ with period $\dfrac{\pi}{\omega}$ (III).
- B: $\sin^3(2\omega t)$ repeats every $\dfrac{\pi}{\omega}$, but it contains two frequencies ($2\omega$ and $6\omega$), so it is not SHM (I).
- C: the frequencies $\omega$ and $\pi\omega$ have an irrational ratio, so the motion never repeats (IV).
- D: periods $\dfrac{2\pi}{\omega}$ and $\dfrac{\pi}{\omega}$ give an overall period of $\dfrac{2\pi}{\omega}$; it is not SHM (II).
Solution by Sreeraj P, M.Sc Physics