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Question

(B) The analysis of the given functions is as follows:
$A$. $\sin^2 \omega t = \frac{1-\cos 2\omega t}{2}$. This is a periodic function with period $

Vedclass · Accepted Answer

$A-II, B-I, C-III, D-IV$

Vedclass · Answer

(B) The analysis of the given functions is as follows:
$A$. $\sin^2 \omega t = \frac{1-\cos 2\omega t}{2}$. This is a periodic function with period $T = \pi/\omega$, but it is not Simple Harmonic Motion $(SHM)$ because it contains a constant term and a frequency $2\omega$. Thus, $A-II$.
$B$. $\sin^3 \omega t = \frac{3\sin \omega t - \sin 3\omega t}{4}$. This is a periodic function with period $T = 2\pi/\omega$, but it is not $SHM$ because it involves multiple frequencies. Thus, $B-I$.
$C$. $\sin \omega t + \cos \pi \omega t$ is non-periodic because the ratio of the frequencies $\omega/\pi\omega = 1/\pi$ is an irrational number. Thus, $C-III$.
$D$. $\cos \omega t + \cos 2\omega t$ is a periodic function with period $T = 2\pi/\omega$, but it is not $SHM$ as it is a superposition of two different frequencies. Thus, $D-IV$.
Therefore, the correct matching is $A-II, B-I, C-III, D-IV$.

List-$I$	List-$II$
$A$. $\sin^2 \omega t$	$I$. Periodic but not $SHM$ $(T = 2\pi/\omega)$
$B$. $\sin^3 \omega t$	$II$. Periodic but not $SHM$ $(T = \pi/\omega)$
$C$. $\sin \omega t + \cos \pi \omega t$	$III$. Non-periodic
$D$. $\cos \omega t + \cos 2\omega t$	$IV$. Periodic but not $SHM$ $(T = 2\pi/\omega)$

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