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
Concept. A motion is SHM only if it can be written as one sinusoid (plus a constant). A sum of sinusoids is periodic only if their frequencies have a rational ratio; the period is then the LCM of the periods.
Step 1. A. $\sin^2\omega t=\tfrac12-\tfrac12\cos2\omega t$: SHM about $\tfrac12$ with $T=\dfrac\pi\omega$. III.
Step 2. B. $\sin^3 2\omega t=\tfrac34\sin2\omega t-\tfrac14\sin6\omega t$: two frequencies, period $\dfrac\pi\omega$, not SHM. I.
Step 3. C. Frequencies $\omega$ and $\pi\omega$: ratio $\pi$ is irrational, so never repeats. IV.
Step 4. D. Frequencies $\omega$ and $2\omega$: period $\dfrac{2\pi}\omega$, not SHM. II.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
