Question

Match the column : \[ \begin{array}{|c|c|} \hline \text{Column-I} & \text{Column-II} \\ \hline (A)\;\sin^2(\omega t) & (P)\; \text{Non-periodic} \\ (B)\;\cos(\omega t)+\cos(2\omega t) & (Q)\; \text{Periodic but not SHM} \\ (C)\;\sin^2(2\omega t) & (R)\; \text{SHM with time period } \frac{\pi}{2\omega} \\ (D)\;\cos(\pi+\omega t)+\cos(\omega t) & (S)\; \text{SHM with time period } \frac{\pi}{\omega} \\ \hline \end{array} \]

• A. A-R, B-P, C-S, D-Q
• B. A-S, B-Q, C-R, D-P
• C. A-R, B-P, C-Q, D-S
• D. A-S, B-Q, C-P, D-R

Hint:

Useful identity: \[ \sin^2 x=\frac{1-\cos 2x}{2} \] This helps convert squared trigonometric functions into cosine functions to determine period.

Answer 1

The Correct Option is B

Concept: For SHM, the displacement must be a single sine or cosine function of time. Functions such as \(\sin^2(\omega t)\) can be rewritten using trigonometric identities to determine periodicity.
Step 1: Analyze (A). \[ \sin^2(\omega t)=\frac{1-\cos(2\omega t)}{2} \] This is periodic and behaves as SHM form with angular frequency \(2\omega\). \[ 2\omega t=2\pi \] \[ T=\frac{\pi}{\omega} \] Thus, \[ A \rightarrow (S) \]
Step 2: Analyze (B). \[ \cos(\omega t)+\cos(2\omega t) \] This is a sum of two cosine functions of different frequencies. Hence it is periodic but not simple harmonic motion. \[ B \rightarrow (Q) \]
Step 3: Analyze (C). \[ \sin^2(2\omega t)=\frac{1-\cos(4\omega t)}{2} \] Angular frequency \(=4\omega\). \[ T=\frac{2\pi}{4\omega}=\frac{\pi}{2\omega} \] Thus, \[ C \rightarrow (R) \]
Step 4: Analyze (D). \[ \cos(\pi+\omega t)=-\cos(\omega t) \] \[ \cos(\pi+\omega t)+\cos(\omega t)=-\cos(\omega t)+\cos(\omega t)=0 \] This is constant and hence non-periodic. \[ D \rightarrow (P) \] Final matching: \[ A-S,\; B-Q,\; C-R,\; D-P \] \[ \boxed{(2)} \]