If the function sin^2 omega t (where t is tim | vedclass

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(B) The given function is $f(t) = \sin^2 \omega t$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we can rewrite the fu

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$\frac{\pi}{\omega} \ s$

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(B) The given function is $f(t) = \sin^2 \omega t$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we can rewrite the function as:$f(t) = \frac{1 - \cos(2\omega t)}{2} = \frac{1}{2} - \frac{1}{2} \cos(2\omega t)$.The term $\cos(2\omega t)$ represents a periodic function with angular frequency $\omega' = 2\omega$.The time period $T$ of a periodic function $\cos(\omega' t)$ is given by $T = \frac{2\pi}{\omega'}$.Substituting $\omega' = 2\omega$,we get $T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$.Thus,the period of the motion is $\frac{\pi}{\omega} \ s$.

If the function $\sin^2 \omega t$ (where $t$ is time in seconds) represents a periodic motion,then the period of the motion is:

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