Q. 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
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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.
