# Is projection of a simple pendulum, doing SHM as well?

I know projection/shadow of a Uniform Cirular Motion does SHM, and a simple pendulum also does shm. But I was wondering whether, for a pendulum in $$xy$$ plane having its central axis parallel to $$y$$ axis, its projection on $$x$$ axis would also do shm on the $$x$$ axis?

A simple pendulum does approximate shm. Suppose that we displace it through angle $$\theta$$ from its equilibrium position. If $$s$$ is its displacement along the arc from its equilibrium position, then the equation $$\ddot s+\omega^2 s=0$$ in which $$\omega=\sqrt {g/l}$$, is only approximately true. The actual equation is $$\ddot \theta+\omega^2 \sin \theta =0\ \ \ \ \ \ \ \ \text {that is}\ \ \ \ \ \ \ \ \ddot s+\omega^2 l\sin (s/l)=0$$ The projection of the pendulum motion on to the $$x$$-axis doesn't make things any better. It is still only approximate shm. If $$x$$ is the displacement from the equilibrium position, then $$\ddot x+g\tan \theta =0\ \ \ \ \ \ \text{that is}\ \ \ \ \ \ \ddot x=\omega^2\frac x {\sqrt {1-(x/l)^2}}.$$

For $$l= 1.00000$$ m and $$\theta =10.000°=0.17453$$ rad, we find $$s=0.1745$$ m, $$l\sin(s/l)=0.1736$$ m, $$\frac x{\sqrt{1-(x/l)^2}}=\tan \theta=0.1763$$ m.

As $$\theta \to 0,\ \ \sin (s/l) \to s/l\ \ \text{and}\ \ \ 1-(x/l)^2 \to 1$$, so both motion along the arc and the $$x$$-axis projection approach shm, because the equations collapse to $$\ddot s+\omega^2 s=0\ \ \ \text{and}\ \ \ \ddot x+\omega^2 x=0$$ .

[The projection of uniform circular motion on to a straight line in the plane of the circle, such as a circle diameter, is true shm. ]

If we have a point moving on a circle of radius $$R$$ with an angular coordinate $$\theta(t)~$$ ($$\theta=0$$ coinciding with the $$y$$ axis), and we project it onto a line perpendicular to the $$y$$ axis, the projection $$x(t)$$ of the circular motion is given by $$x(t) = R ~sin(\theta(t)).$$

Therefore, a SHM $$\theta(t)=A~cos(\omega t + \phi)$$ will correspond to a projected motion $$x(t)=R~sin(A~cos(\omega t + \phi)).$$

This is a SHM only in the limit of a small amplitude of oscillations. However, a simple pendulum behaves like a SHM at this same limit.

For large amplitude motions, not only does the angle of deviation from the vertical cease to behave like a SHM (the period increases, and the intervals of low speed at the extremes of oscillation increase), but also a qualitative difference between the angular evolution and its projection appears as soon as the extreme angles of oscillation become larger than $$\pm\pi/2$$. Indeed, in such cases, a retrograde motion of the projection appears. An example of such a situation is shown in the following figure.