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

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(C) The function $f(t) = \sin^2(\omega t)$ can be written as $f(t) = \frac{1 - \cos(2\omega t)}{2} = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$.This is

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a periodic,but not simple harmonic motion with a period $\frac{\pi}{\omega}$

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(C) The function $f(t) = \sin^2(\omega t)$ can be written as $f(t) = \frac{1 - \cos(2\omega t)}{2} = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$.This is a periodic function because it is a sum of a constant and a cosine function.The angular frequency of this function is $2\omega$. Therefore,the time period $T$ is given by $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.For a motion to be Simple Harmonic Motion $(SHM)$,the acceleration must be proportional to the negative of the displacement,i.e.,$a = \frac{d^2y}{dt^2} \propto -y$.Let $y = \sin^2(\omega t)$.Then,$\frac{dy}{dt} = 2\sin(\omega t) \cdot \cos(\omega t) \cdot \omega = \omega \sin(2\omega t)$.And,$\frac{d^2y}{dt^2} = \omega \cdot \cos(2\omega t) \cdot 2\omega = 2\omega^2 \cos(2\omega t)$.Since $\frac{d^2y}{dt^2}$ is not proportional to $-y$ (because $\cos(2\omega t)$ is not proportional to $\sin^2(\omega t)$),the motion is periodic but not $SHM$.Thus,the function represents a periodic,but not simple harmonic motion with a period $\frac{\pi}{\omega}$.

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

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