Motion of a particle in x-y plane is describe | vedclass

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(A) Given equations are $x = 4 \sin \left(\frac{\pi}{2} - \omega t\right)$ and $y = 4 \sin (\omega t)$.Using the trigonometric identity $\sin \left(\f

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circular

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(A) Given equations are $x = 4 \sin \left(\frac{\pi}{2} - \omega tight)$ and $y = 4 \sin (\omega t)$.Using the trigonometric identity $\sin \left(\frac{\pi}{2} - 	hetaight) = \cos 	heta$,we can rewrite the equation for $x$ as $x = 4 \cos (\omega t)$.Now we have $x = 4 \cos (\omega t)$ and $y = 4 \sin (\omega t)$.To find the path,square and add both equations:$x^2 + y^2 = (4 \cos \omega t)^2 + (4 \sin \omega t)^2$$x^2 + y^2 = 16 \cos^2 \omega t + 16 \sin^2 \omega t$$x^2 + y^2 = 16 (\cos^2 \omega t + \sin^2 \omega t)$Since $\cos^2 	heta + \sin^2 	heta = 1$,we get $x^2 + y^2 = 4^2$.This is the equation of a circle with radius $4 	ext{ m}$ centered at the origin. Therefore,the path of the particle is circular.

Motion of a particle in $x-y$ plane is described by a set of following equations $x=4 \sin \left(\frac{\pi}{2}-\omega t\right) \text{ m}$ and $y=4 \sin (\omega t) \text{ m}$. The path of the particle will be:

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