Show bigger amplitude of physical pendulum means bigger period Suppose you have a physical pendulum. It is true that as amplitude increases, the period increases. Can we demonstrate this fact without explicitly finding the period (which is pretty involved and pretty messy) in: 


*

*an intuitive fashion, 

*rigorously?
 A: Let's draw our pendulum:

The equation of motion is:
$$ F\ell = -I\frac{d^2\theta}{dt^2} $$
This may seem a bit unfamiliar, but it's just the circular motion equivalent of $F = ma$. We replace the force by the torque, $F\ell$, the mass by the moment of inertia $I$ and the acceleration by the angular acceleration $\ddot{\theta}$. A bit of quick geometry gives us $F = mg\sin\theta$, so our equation becomes:
$$ mg\sin\theta\ell = -I\frac{d^2\theta}{dt^2} $$
Assuming our mass is a point, the moment of intertia is just $I = m\ell^2$, and with a quick rearrangement we get:
$$ \frac{d^2\theta}{dt^2} = -\frac{g}{\ell}\sin\theta $$
What your physics teacher will do next is point out that $\sin\theta$ can be expanded as a power series:
$$ \sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - ... $$
and if $\theta$ is small then the higher powers of $\theta$ are very small and we get $\sin\theta \approx \theta$. Substitute this for $\sin\theta$ in our equation above and we get:
$$ \frac{d^2\theta}{dt^2} = -\frac{g}{\ell}\theta \tag{1} $$
which is our good old simple harmonic motion equation.
Now we can answer your question, because if we keep increasing the angle of swing we're going to get to a point where the $\theta^3$ term is too large to be ignored. In that case our equation (1) becomes:
$$ \frac{d^2\theta}{dt^2} = -\frac{g}{\ell}\left(\theta - \frac{\theta^3}{3!}\right) \tag{2} $$
Now take two pendulums (penduli?), one described by the simple harmonic equation (1) and one described my our more accurate equation (2), and start them at some initial angle $\theta_0$. The angular acceleration calculated by equation (2) is less than the angular acceleration calculated by equation (1) for all values of $\theta$ (except at $\theta = 0$). So if both penduli start at the same place, $\theta_0$, pendulum 2 must take longer to get to $\theta = 0$ than pendulum 1 will. But this time is just a quarter of the period, and that means the period of pendulum 2 must be greater than the period of pendulum 1. So for a real pendulum the period must increase with increasing amplitude of swing.
A: Here's a solution which is similar to John Rennie's but hopefully less involved.  I'll steal his image, too:

The pendulum has kinetic energy $T$, potential energy $U$, and total energy $E=T+U$, where
$$ T = \frac12 m\ell^2 \dot\theta^2, \quad\quad U = mg\ell(1-\cos\theta). $$
The simple harmonic approximation takes the limit $\theta\ll1$, where
$$
U = mg\ell \left( \frac{\theta^2}{2!} - \frac{\theta^2}{4!} + \cdots \right)
\approx mg\ell \frac{\theta^2}2 \equiv U_\text{quadratic}
$$
Now it's clear at small $\theta$, and happens to be the case for all $\theta$, that $U_\text{quadratic}$ is an overestimate of $U$:

Therefore whatever our starting $\theta$ happens to be, using the simple harmonic approximation $U_\text{quadratic}$ predicts too much total energy $E$, and correspondingly too much kinetic energy $T$ --- our physical pendulum goes slower than in the approximation.  The prediction of a constant period $\tau_\text{quadratic} = 2\pi\sqrt{\ell/g}$ is therefore an underestimate, and the underestimate gets worse for large amplitudes, so the period must increase with amplitude.
A: The frequency of a simple pendulum is easy to compute, even for large angular amplitudes. Consider the pendulum in the figure below.
For small amplitudes ($2l-d << l$) this pendulum oscillates at angular frequency $\omega_+=\sqrt{g/l}$. 
Now we define a pendulum with a 'reciprocal' length $l'=2l^2/d$. This reciprocal pendulum has a small amplitude frequency $\omega_-=\sqrt{gd/2l^2}$.
Quite amazingly, the frequency of the original pendulum (of length $l$) can be written as $$\omega=agm(\omega_+,\omega_-)$$
Here, $agm$ denotes the arithmetic-geometric mean, first studied by Gauss.
Now, it is easy to see that based on $d<2l$ the arm of the reciprocal pendulum is always longer than the arm of the original pendulum, and therefore $\omega_-<\omega_+$ always holds, and $\omega_-$ decreases whenever $d$ decreases. The $agm$ is real easy to compute (Google is your friend here), but all we need to ensure that $\omega$ decreases when $d$ decreases (and hence the amplitude increases), is the simple fact that $agm$ really acts as a mean. In other words, $agm(\omega_+,\omega_-)$ decreases if $\omega_-$ decreases while $\omega_+$ stays fixed. Presto.
A: The risk of an "intuitive" answer is that people's intuitions often go awry.
Here's an intuitive answer: Take two identical pendulums.  Release one at $\theta$ and the other at $2*\theta$ from vertical.  No matter what else, the pendulum that starts from farther away can never "catch up" to the other one.  In addition, since it's obvious that it will have greater K.E. at vertical, it will travel farther "up" the far side.   After all, if you drop two balls from different heights, the higher one will always take longer to reach the ground. (except now you need to prove that, too :-( ) 
That's great, but someone else will say "but what if the $2*\theta$ pendulum accelerates so fast it catches up?" [sort of like the infamous "hot water freezes faster than cold" argument]   And so on.   At some point you're going to have to use a little math to prove your case. 
