#### Question

Show that S.H.M. is a projection of U.C.M. on any diameter

#### Solution

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.

`cos(omegat+alpha)=x/a`

`thereforex=acos(omegat+alpha)` .........(Equ.1)

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

`v=dx/dt=d/dt[acos(omegat+alpha)]`

`therefore v=-aomegasin(omegat+alpha)` .............(Equ.2)

As acceleration of particle is the time rate of change of velocity, we have

`a=(dv)/dt=d/dt [-aomegasin(omegat+alpha)]`

`thereforea=-aomega^2cos(omegat+alpha)`

`therefore a=-omega^2x`

Hence, the projection of a uniform circular motion on a diameter of a circle is simple harmonic motion.