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Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter.

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Consider a particle which moves anticlockwise around a circular path of radius A with a constant angular speed ω. Let the path lie in the x-y plane with the centre at the origin O. The instantaneous position P of the particle is called the reference point and the circle in which the particle moves as the reference circle.

The perpendicular projection of P onto the yaxis is Q. Then, as the particle travels around the circle, Q moves to-and-fro along the yaxis. Line OP makes an angle α with the x-axis at t = 0. At time t, this angle becomes θ = ωt + α.

The projection Q of the reference point is described by the y-coordinate,

y = OQ = OP sin ∠OPQ, Since ∠OPQ = ωt + α, y = A sin(ωt + α) which is the equation of a linear SHM of amplitude A. The angular frequency w of a linear SHM can thus be understood as the angular velocity of the reference particle.

The tangential velocity of the reference particle is v = ωA. Its y-component at time t is vy = ωA sin (90° – θ) = ωA cos θ 

∴ vy = ωA cos (ωt + α)

The centripetal acceleration of the reference particle is ar = ω2A, so that its y-component at time t is ax = ar sin ∠OPQ 

∴ ax = – ω2 A sin (ωt + α)

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