Two simple harmonic motions given by, `x = a sin (omega t+delta)` and `y = a sin (omega t + delta + (pi)/(2))` act on a particle will be

Options

A. circular anti-clockwise
B. elliptical anti-clockwise
C. elliptical clockwise
D. circular clockwise

Answer

D

Explanation

Given, `x = A sin (omega t + delta)" "...(i)`
and `y = A sin(omega t + delta + (pi)/(2)) = A cos (omega t + delta)" "...(ii)`
Squaring and adding Eqs. (i) and (ii), we get
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) = (pi)/(2), x = A, y = 0`
At `(omega t + delta) = pi, x = 0, y = - A`
At `(omega t + delta) = (f 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.