Question

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

Answer 1

Answer: (D)

circular clockwise

Answer 2

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.