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)

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.