A) circular anti-clockwise
B) elliptical anti-clockwise
C) elliptical clockwise
D) circular clockwise
Correct Answer: D
Solution :
Given, \[x=A\sin (\omega t+\delta )\] ... (i) and \[y=A\sin \left( \omega t+\delta +\frac{\pi }{2} \right)\] \[=A\cos (\omega t+\delta )\] ... (ii) Squaring and adding Eqs. (i) and (ii), we get \[{{x}^{2}}+{{y}^{2}}={{A}^{2}}[{{\sin }^{2}}(\omega t+\delta )+{{\cos }^{2}}(\omega t+\delta )]\] Or \[{{x}^{2}}+{{y}^{2}}={{A}^{2}}\] which is the equation of a circle. Now, At\[(\omega t+\delta )=0,\,\,x=0,\,\,y=0\] At \[(\omega t+\delta )=\frac{\pi }{2},\,\,x=A,\,\,y=0\] At \[(\omega t+\delta )=\pi ,\,\,x=0,\,\,y=-A\] At \[(\omega t+\delta )=\frac{3\pi }{2},\,\,x=-A,\,\,y=0\] At \[(\omega t+\delta )=2\pi ,\,\,x=A,\,\,y=0\] From the above data, the motion of a particle is a circle transversed in clockwise direction.You need to login to perform this action.
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