A) circular anti-clockwise
B) elliptical anti-clockwise
C) elliptical clockwise
D) circular clockwise
Correct Answer: D
Solution :
(d )Given, \[x=A\,\sin (\omega t+\delta )\] ...(i) and \[y=A\,\,\sin \left( \omega t+\delta +\frac{\pi }{2} \right)\] \[=A\,\cos (\omega \tau +\delta )\] ?(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 \[(\omega t+\delta )=0,\text{ }x=0,\text{ }y=0\] 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.
You will be redirected in
3 sec