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# SHM as a Projection of Circular Motion

The particle is at position P at t = 0 and revolves with a constant angular velocity (ω) along a circle. The projection of P on the diameter along the x-axis (M). At the later time (t) the particle is at Q. Now its projection on the diameter along the x-axis is N.

As the particle P revolves around in a circle anti-clockwise its projection M follows it up moving back and forth along the diameter such that the displacement of the point of projection at any time (t) is the x-component of the radius vector (A).

x = A cos ( ωt + Φ) . . . . . . . (1)

y = A sin ( ωt + Φ) . . . . . (2)

Thus, we see that the uniform circular motion is the combination of two mutually perpendicular linear harmonic oscillation.

It implies that P is under uniform circular motion, (M and N) and (K and L) are performing simple harmonic motion about O with the same angular speed ω as that of P.

P is under uniform circular motion, which will have centripetal acceleration along A (radius vector)\overrightarrow{{{a}_{c}}}=A{{\omega }^{2}}ac​​=Aω2 (towards the centre)

It can be resolved into two components:

• aN​=2sin2(ωt+ϕ)
• aL​=2cos2(ωt+ϕ)

aN and aL acceleration corresponding to the points N and L respectively.

In the above discussion, the foot of projection on the x-axis is called horizontal phasor.

Similarly, the foot of the perpendicular on the y-axis is called vertical phasor. Already we know the vertical and horizontal phasor will execute the simple harmonic motion of amplitude A and angular frequency ω. The phases of the two SHM differ by π/2.