question_answer

Two simple harmonic motions are given by \[y=A\sin \left( \omega t+\delta  \right)\]and \[y=A\sin \left( \omega t+\delta +\frac{\pi }{2} \right)\] act on a panicle 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 :

                (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.