^{-1}).

Here in the figure

OQ→position of Q relative to O

θ→angle OQ vector makes with the positive x-axis

This vector whose horizontal component represents the actual motion is called a

**phasor**

Let P be the point representing the projection of point Q (known as

*) on the horizontal diameter of the circle (known as*

**the reference point***). As the reference point revolves, the point P moves back and forth along the horizontal diameter always keeping below or above the point Q. The motion of this point P is comparable to that of a body moving under the influence of elastic restoring force and executing SHM in the absence of frictional forces.*

**reference circle**From the figure given above, displacement of point P at any given time t is the distance OP or x and from the figure

x=Acosθ

The angular velocity of circular motion is

$\omega =\frac{\mathrm{angle}\mathrm{swept}}{\mathrm{time-taken}}=\frac{\theta}{t}$

so,

θ=ωt

and

x=Acosωt (1)

The figure given below shows the velocity in SHM using the circle of reference.

Velocity of the reference point Q would be the component of Q’s velocity parallel to the x-axis since P is always directed below or above the reference point. So,

v_{P}=-vsinθ = -ωAsinωt (2)

since v=ωA is the tangential velocity of motion of reference point Q. The negative sign is introduced because the direction of velocity is towards the left. When Q is below horizontal diameter, the velocity of P is towards the right, but since sinθ is negative at such points, the minus sign is still needed.

Figure given below shows the acceleration in SHM using circle of reference.

The acceleration of Q on reference circle is directed radially inwards towards O as can be seen in the figure and its magnitude is equal to

From figure x-component of acceleration is acceleration of point P. So,

a_{P}=-acosθ= -ω^{2}Acosθ (3)

The minus sign is introduced because the acceleration is towards the left. When Q is to the left of the center, the acceleration of P is towards the right but, since cosθ is negative at such points, the minus sign is still needed.

To prove that motion of point P is simple harmonic, we now use equation 1 in equation 3 then we have

a_{P}=-ω^{2}x

since ω is a constant, the acceleration of point P at any instant is equal to negative constant times displacement xat that instant and this is just the essential feature of SHM and this proves that motion of point P is indeed SIMPLE HARMONIC.