question_answer

                Two simple harmonic motions given by \[x=A\sin (\omega t+\delta )\] and \[y=A\,\sin \left( \omega t+\delta +\frac{\pi }{2} \right)\] act on a particle simultaneously; then the motion of particle will be:

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.