A) circular anti-clockwise
B) circular clockwise
C) elliptical anti-clockwise
D) elliptical clockwise
Correct Answer: B
Solution :
Two simple harmonic motions be written as |
\[x=A\sin (\omega t+\delta )\] (i) |
and \[y=A\,\,\sin \,\left( \omega t+\delta +\frac{\pi }{2} \right)\] |
or \[y=A\cos (\omega t+\delta )\] (ii) |
Squaring and adding Eqs. (i) and (ii) we obtain |
\[{{x}^{2}}+{{y}^{2}}={{A}^{2}}\,[\sin (\omega t+\delta )+{{\cos }^{2}}(\omega t+\delta )]\] |
or \[{{x}^{2}}+{{y}^{2}}={{A}^{2}}\] \[(\because {{\sin }^{2}}\,\theta +{{\cos }^{2}}\,\theta =1)\] |
This is the equation of a circle. |
At \[\,\omega t+\delta =0\,;\,x=0,\,y=A\] |
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 \[\omega t+\delta =2\pi \,\,\,;\,x=0,\,y=A\] |
Thus, it is obvious that motion of particle is traversed in clockwise direction. |
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