The function sin^2(omega t) represents: | vedclass

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(D) Given function is $y = \sin^2(\omega t)$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos(2	heta)}{2}$,we can rewrite the functio

Vedclass · Accepted Answer

$A$ periodic but not simple harmonic motion with a period $\pi /\omega $

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(D) Given function is $y = \sin^2(\omega t)$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos(2	heta)}{2}$,we can rewrite the function as:$y = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$.The period $T$ of a function $\cos(kt)$ is given by $T = \frac{2\pi}{k}$.Here,$k = 2\omega$,so the period is $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.Simple Harmonic Motion $(S.H.M.)$ must satisfy the differential equation $\frac{d^2y}{dt^2} = -\Omega^2 y$. The given function represents a constant offset plus a cosine term,which does not satisfy the condition for $S.H.M.$ because the equilibrium position is shifted and it is not a pure sinusoidal oscillation about the origin. Therefore,it is a periodic motion but not $S.H.M.$

The function $\sin^2(\omega t)$ represents:

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