Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

From the given functions, identify the function which represents a periodic motion:

$e^{\omega t}$

$\log_e(\omega t)$

$\sin \omega t + \cos \omega t$

$e^{-\omega t}$

Definition of Periodic Motion: A motion that repeats itself at regular intervals of time is called periodic motion. Mathematically, a function $f(t)$ is periodic if $f(t) = f(t + T)$ for some period $T$ .
Analysis of Options:

$e^{\omega t}$ : This is an exponential growth function. It increases monotonically with time and never repeats. Hence, it is non-periodic.
$\log_e(\omega t)$ : This is a logarithmic function. It increases monotonically with time and never repeats. Hence, it is non-periodic.
$e^{-\omega t}$ : This is an exponential decay function. It decreases monotonically towards zero but never repeats. Hence, it is non-periodic.
$\sin \omega t + \cos \omega t$ : Both sine and cosine functions are periodic with period $T = 2\pi/\omega$ . The sum of two periodic functions with the same frequency is also periodic. We can rewrite this expression: $y = \sin \omega t + \cos \omega t = \sqrt{2} \left( \frac{1}{\sqrt{2}}\sin \omega t + \frac{1}{\sqrt{2}}\cos \omega t \right)$ $y = \sqrt{2} \left( \sin \omega t \cos \frac{\pi}{4} + \cos \omega t \sin \frac{\pi}{4} \right) = \sqrt{2} \sin(\omega t + \frac{\pi}{4})$ This represents a Simple Harmonic Motion (a specific type of periodic motion) with period $T = 2\pi/\omega$ . Hence, it is periodic.

Conclusion: Only the function involving sine and cosine represents periodic motion.

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