Question

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

Answer 1

Answer: (D)

(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 $(\omega t+\delta )=0,x=0,y=0$ 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.