Question

Two simple harmonic motions given by $x=A\sin (\omega t+\delta )$ and $y=A\sin (\omega t+\delta +\frac{\pi }{2})$ 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

The correct option is D circular clockwise

Given, $x=A\sin (\omega t+\delta )$ ...(i)
and $y=A\sin (\omega t+\delta +\frac{\pi }{2})$
$=A\cos (\omega t+\delta )$ ...(ii)


Squaring and adding Eqs. (i) and (ii), we get
$x^2+y^2=A^2\left[{\sin }^2\right(\omega t+\delta )+{\cos }^2(\omega t+\delta \left)\right]$
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, we can see that the motion of the particle is a circle transversed in clockwise direction.
Hence, (d) is the correct answer.