A) straight line
B) circle
C) ellipse
D) figure of 8
Correct Answer: C
Solution :
\[x=A\sin \omega t\] \[{{x}^{'}}=B\,\sin (wt\,+\pi /2)\,=B\cos \omega t\] \[{{\sin }^{2}}\omega t\,+{{\cos }^{2}}\omega t\,=\frac{{{x}^{2}}}{{{A}^{2}}}\,+\frac{{{x}^{'2}}}{{{B}^{2}}}=1\] hence ellipse.You need to login to perform this action.
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