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### Circle of reference in Simple Harmonic Motion

Circle of reference is a graphical representation which facilitate the understanding of various SHM relationships. Consider the figure given below in which a point Q moves anti-clockwise around a circle of radius A with a constant angular velocity ω (in rad.s-1).
Here in the figure
OQ→position of Q relative to O
θ→angle OQ vector makes with 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 reference point) on the horizontal diameter of the circle (known as reference circle). As the reference point revolvs , 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 elstic restoring force and executing SHM in the absence of frictional forces.
From 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
$\omega =\frac{\mathrm{angle}\mathrm{swept}}{\mathrm{timetaken}}=\frac{\theta }{t}$
so,
θ=ωt
and
x=Acosωt                                               (1)
Figure given below shows the velocity in SHM using circle of reference.
Velociy 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,
vP=-vsinθ = -ωAsinωt                            (2)
since v=ωA is the tangential velocity of motion of reference point Q. The negative sign is introduced because 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
$a=\frac{{v}^{2}}{A}={\omega }^{2}A$
From figure x-component of acceleration is acceleration of point P. So,
aP=-acosθ= -ω2Acosθ                           (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
aP=-ω2x
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 essential feature of SHM and this proves that motion of point P is indeed SIMPLE HARMONIC.