question_answer

Two simple harmonic motions are 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)  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.