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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?

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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. ]

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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. enter image description here

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