Show that S.H.M. is a projection of U.C.M. on any diameter
Suppose that particle P starts from the initial position with initial phase α (angle between radius OP and the x-axis at the time t = 0).
In time t, the angle between OP and x-axis is (ωt +α) as particle P moving with constant angular velocity (ω) as shown.
This is the expression for displacement of particle M at time t
As velocity of the particle is the time rate of change of displacement then we have
`therefore v=-aomegasin(omegat+alpha)` .............(Equ.2)
As acceleration of particle is the time rate of change of velocity, we have
Hence, the projection of a uniform circular motion on a diameter of a circle is simple harmonic motion.