Question

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

Answer 1

Let us consider a particle P be in a uniform circular motion about O in the xy plane.
Let OP = r
then, x = $rcos\theta$
y = $rsin\theta$
here motion is inform circular motion then, $\theta$ increase in a constant rate.
e.g., $\frac{\theta}{t}=\omega$
so, $\theta = \omega t$ ....(1)

put equation (1) in x = $rcos\theta$
y = $rsin\theta$
then, x = $rcos\omega t$.....(2)
y = $rsin\omega t$

after double differentiation of equation (2),
we get, d²x/dt² = $-\omega^2rcos\omega t=-\omega^2x$
we know, d²x/dt² = a [ linear acceleration along x axis ]
so, F = ma = -$m\omega^2x$
if we assume , K = $m\omega^2$
then, F = -Kx

this force is directly proportional to the displacement. then the motion is simple harmonic motion. so, projection of uniform circular motion on any diameter is in linear SHM.