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

Step 1: Understanding the Concept:
To classify these mathematical functions, we analyze their trigonometric expansions.
Simple Harmonic Motion (SHM) is characterized by a single sine or cosine function with a definite frequency.
The superposition of harmonically related frequencies usually results in periodic motion but not pure SHM.
Step 2: Key Formula or Approach:
Use the trigonometric identities $\sin^2\theta = \frac{1-\cos 2\theta}{2}$ and $\cos(\pi+\theta) = -\cos\theta$ to simplify the expressions to determine the time period $T = \frac{2\pi}{\omega_{\text{eq}}}$.
Step 3: Detailed Explanation:
For (A): $\sin^2 \omega t$
Using the identity $\cos 2\theta = 1 - 2\sin^2 \theta$:
\[ \sin^2 \omega t = \frac{1 - \cos 2\omega t}{2} = \frac{1}{2} - \frac{1}{2} \cos 2\omega t \]
This equation represents SHM oscillating around a mean position of $1/2$ with an angular frequency of $2\omega$.
The time period $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$. Matches (S).
For (B): $\cos (\omega t) + \cos (2\omega t)$
This is a superposition of two cosine waves with commensurable angular frequencies $\omega$ and $2\omega$.
Since there is more than one frequency component, it does not represent pure SHM.
However, it is periodic with the fundamental period being the LCM of individual periods. The period is $T = \frac{2\pi}{\omega}$.
Matches (Q).
For (C): $\sin^2 (2\omega t)$
Using the same identity as in (A):
\[ \sin^2 (2\omega t) = \frac{1 - \cos 4\omega t}{2} = \frac{1}{2} - \frac{1}{2} \cos 4\omega t \]
This equation represents SHM with an angular frequency of $4\omega$.
The time period $T = \frac{2\pi}{4\omega} = \frac{\pi}{2\omega}$. Matches (R).
For (D): $\cos (\pi + \omega t) + \cos (\omega t)$
Using the identity $\cos (\pi + \theta) = -\cos \theta$:
\[ -\cos \omega t + \cos \omega t = 0 \]
Since the function evaluates to exactly zero identically, it exhibits no oscillation and is non-periodic. Matches (P).
Step 4: Final Answer:
The matching is A-S, B-Q, C-R, D-P.