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I'm looking for an intuition on the relationship between time period and amplitude (for a large pertubation) of pendulums. Why does the period depend on the amplitude? I know the math of the problem. I am looking for physical intuition.

Edit: large perturbations not small perturbations

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    $\begingroup$ Do you mean for large perturbations? I thought period was unaffected by amplitude for small perturbations, no? $\endgroup$ – Tom B. Feb 19 '18 at 4:41
  • $\begingroup$ Check out physics.stackexchange.com/questions/346683/… $\endgroup$ – jim Feb 20 '18 at 9:40
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For simple harmonic motion (that is, oscillatory motion for which the period is independent of amplitude), the restoring force must be proportional to the displacement. If gravity is the restoring force, as for a bead sliding frictionlessly on an upward-curved wire, then the angle of inclination of the wire must change as you move along it in such a way that the component of gravity acting along the wire, $g \sin\theta$, increases linearly with distance traveled along the wire. If you work out this shape, (see tautochrone) it is a cycloid, which starts out pretty flat, and curves upward more and more sharply. This is, then, the shape that a pendulum bob must follow in order to execute simple harmonic motion. In a simple pendulum, on the other hand, the bob follows a circular arc (constant curvature). Since the curvature does not increase with distance, as it must for constant period, the period must increase with amplitude.

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If the pendulum is standing upright vertically (at $\pi$) with zero velocity then the period is infinity since (assuming it's perfectly balanced) it will stay upright forever.

Now make the amplitude of the pendulum oscillations close to $\pi$, so that it nearly goes vertical and velocity is small there, making it stay sticking upright for a long time as the situation is close to the first scenario. This obviously effects the period, but only happens when the amplitude is large enough.

Edit: Just noticed you said for small perturbation. Like the other comment said, the period is a constant for small perturbations.

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Dimensional analysis may help. Since the period is measured in seconds, and the problem involves mass $M$, gravity $g$ and the length $\ell$ of the pendulum in addition to the angle, one may ask what combination of these variable will have the dimension of time. The answer is simply $$ T=\sqrt{\frac{\ell}{g}}\,f(\theta_0) $$ where $f$ is a function of the (dimensionless) variable $\theta_0$, the amplitude of the motion. No other combination of the variables will produce something with units of time.

The function $f$ must be determined by experiment or by a more detailed model than just dimensional analysis. Indeed, one can show that $$ f\approx 2\pi\left(1+\frac{\theta_0^2}{16}\right) $$ For small angles - say $15^\circ= \frac{\pi}{12}$rad - this correction is quite small so, for small angles, the period is largely independent of the amplitude of the oscillation.

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For an object to undergo simple harmonic motion, the amplitude of the restoring force has to get larger with increased amplitude, to give it greater acceleration so it can "catch up" to an object that is started with a smaller amplitude. Specifically, it has to increase as the amplitude squared, but that's not directly applicable for an intuitive argument.

With a pendulum, the restoring force increases with amplitude up to an angle of $90^\circ$, but then it decreases. So it's obvious at least that the period must get larger once the amplitude passes $90^\circ$. And since the restoring force changes smoothly, it shouldn't be surprising that the period doesn't suddenly start increasing at $90^\circ$, but that it instead occurs gradually as the angle gets larger.

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