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.