For a pendulum's motion to be simple harmonic motion (S.H.M.) is it necessary for a pendulum to have small amplitude or S.H.M. can be produced at large amplitudes as well?
If it is really necessary for an S.H.M. to have small amplitudes then why is it? because even at large amplitudes there is restoring force pulling the pendulum toward mean position and its acceleration is directly proportional to the displacement.
[Physics] Pendulum’s motion is simple harmonic motion
accelerationdisplacementharmonic-oscillatornewtonian-mechanicsoscillators
Related Solutions
In the simplest example the concept of phase angle is a convenient way of comparing the motion of two simple harmonic oscillations of the same frequency. $\omega = \frac {2 \pi}{T}$ where $T$ is the period of the oscillation.
If you have two oscillations an oscillation $A$ has a maximum displacement at time $t_A$ and oscillation $B$ reaches a maximum displacement at a time $t_B$ then the phase angle $\phi_{BA}$ can be said to be $ \dfrac {t_B-t_A}{T} \cdot 2 \pi$ where $T$ is the period of the motion. This is the phase of $B$ relative to $A$.
However one could equally say that the phase of $A$ relative to $B$ is what is required then $\phi_{AB} = -\phi_{BA}$.
But then since the motion is continuous you can have other time differences when the motion $A$ is at the same relative position to that of the motion $B$.
The one could write $ \phi_{BA} =\dfrac {(t_B+nT)-t_A}{T} \cdot 2 \pi = \left (\dfrac {t_B-t_A}{T} + n\right)\cdot 2 \pi$ where $n$ is an integer. This represents motion $B$ being many periods and a little bit behind that of motion $A$.
For convenience the phase angle is restricted to the ranges $0\le \phi \le \pi$ or $-\frac \pi 2 \le \phi \le +\frac \pi 2$
The same sort of analysis is true for the motions at different positions and then the period $T$ would be replaced by the wavelength $\lambda$.
Why would you consider the amplitude to be equal to the horizontal displacement? There's also a vertical displacement, which is arguably more important, without vertical displacement there would be no oscillation, since the potential energy would remain the same. Or take for example a torsion pendulum:
You wouldn't describe the amplitude as $r\sin{\theta}$, because that would give an amplitude of zero when the pendulum swings 360 degrees. It makes more sense to say that the amplitude is equal to the length of the path the pendulum travels, i.e. the length of the arc.
Best Answer
As you stated, in order to have simple harmonic motion, you need to have an acceleration that is proportional to the displacement. For a pendulum, if you work out the details, you will find that
$\frac{d^2\theta}{dt^2} \propto -\sin(\theta)$
where $\theta$ is the angle the pendulum makes with the vertical. For small angles, $\sin(\theta)\sim\theta$, which would then lead to simple harmonic motion. For large angles, this approximation no longer holds, and the motion is not considered to be simple harmonic motion.