Question

The phase angle between the projections of uniform circular motion on two mutually perpendicular diameters is
A) $\pi $
B) $\dfrac {3\pi} {4} $
C) $\dfrac {\pi} {2} $
D) Zero

Answer 1

Hint: Here, phase angle is defined as the difference in angle of two bodies moving in a circular motion. Firstly, one body will start moving from one point and finally comes to that point so, while blowing a light on its starting point (i.e. on end of diameter) and at other end of diameter we see a linear motion of projection of body on its horizontal diameter and in same way it is for second body.

Complete answer:
Let us consider two bodies as A and B. A gives its projection on vertical diameter and it looks like it’s moving from right to left and B gives projection on vertical diameter in which it appears to move from down to up. So, as the projection seems to be on diameter only that-is-why, there is a phase difference of $\dfrac {\pi} {2} $.
One projection will be in a mean position. So, we can represent it as –
$x=A\sin \omega t$ ……………………… (1)
And, the other projection is at max position. So, it can be represented as –
$y=A\cos \omega t=A\sin (\omega t+\dfrac {\pi} {2}) $ ……………………. (2)
So, from equation (1) $\Rightarrow $ (2), we get:
Phase difference =$\dfrac {\pi} {2} $

Hence, option (C) is the correct answer.

Note:
We must be knowing the projectile motion of an object in circular motion. Suppose, body A starts moving in circular motion from $0{} ^\circ $ and comes $0{} ^\circ $finally, then it will have its projection at horizontal diameter whereas, body B moves from$90{} ^\circ $ and finally comes to $90{} ^\circ $then, it will have its projection along vertical diameter. So, When two body’s move then, there are two projections created on diameter (one is vertical diameter and, other is horizontal diameter), the body’s projection on diameter moves to and fro.