The displacement of a particle along the x-ax | vedclass

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(A) The given displacement is $x = a \sin^2 \omega t$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we can rewrite the

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simple harmonic motion of frequency $\frac{\omega}{\pi}$

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(A) The given displacement is $x = a \sin^2 \omega t$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we can rewrite the equation as:$x = a \left( \frac{1 - \cos 2\omega t}{2} ight) = \frac{a}{2} - \frac{a}{2} \cos 2\omega t$.This equation represents a periodic motion about the mean position $x = \frac{a}{2}$.For a motion to be Simple Harmonic Motion $(SHM)$,the acceleration must be proportional to the negative of the displacement from the mean position,i.e.,$a_{acc} \propto -(x - x_{mean})$.Here,the displacement is $x - \frac{a}{2} = -\frac{a}{2} \cos 2\omega t$.The velocity is $v = \frac{dx}{dt} = \frac{d}{dt} (\frac{a}{2} - \frac{a}{2} \cos 2\omega t) = a\omega \sin 2\omega t$.The acceleration is $a_{acc} = \frac{dv}{dt} = 2a\omega^2 \cos 2\omega t$.Substituting the expression for displacement,we get $a_{acc} = -4\omega^2 (x - \frac{a}{2})$.Since the acceleration is proportional to the negative of the displacement from the mean position,the motion is $SHM$.The angular frequency of this $SHM$ is $\omega' = 2\omega$.The frequency $f$ is given by $f = \frac{\omega'}{2\pi} = \frac{2\omega}{2\pi} = \frac{\omega}{\pi}$.

The displacement of a particle along the $x$-axis is given by $x = a \sin^2 \omega t$. The motion of the particle corresponds to:

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