Why do large angular displacements affect the amplitude of a simple pendulum? [duplicate]

Question

I was conducting an experiment involving the effect of large angular displacements (10°, 20°, ... 90°) on the amplitude of a simple pendulum. So far, I have found that the period of the pendulum varies according this equation:

\begin{align}T&= 2π\sqrt{\frac {l}{g}} \left(1 + \frac{1}{16} θ^2 + \frac{11}{3072} θ^4 + \frac{173}{737280} θ^6 + \frac{22931}{1321205760} θ^8\right. \\ &\left.\qquad\qquad+ \frac{1319183}{951268147200} θ^{10} + \frac{233526463}{2009078326886400} θ^{12} + …\right) \end{align}

Essentially, period increases with increasing angular displacement. However, I don't really understand why the period varies in this way. Could someone explain this to me without the use of math (no small angle approximations/Taylor series)?

For example, in this way: Although the pendulum accelerates for a longer period of time and has a greater maximum velocity, the additional distance it has to travel is greater than what its larger acceleration and maximum speed can make up for (this is just my guess).

Answer 1

When solving the diferential equation of the motion of a pendulum, to get the solution of $T=2\pi\sqrt{\frac{L}{g}}$, the approximation is used that $sin(x)\approx x$. So that formula becomes less precise when the angles involved become large.

To get a feeling for this:

If $x$ is the length of the arc the pendulum has made from the lowest point, then the angle the string makes with a vertical line is also $x$.

The force that is accelerating the pendulum allong the arc is given by $F=W sin(x)$, with $W$ the weight of the pendulum.

If x is small we can approximate this as $F=Wx$. This is just like the force of a spring that is extended by $x$, with spring constant W. But if x becomes large then $F = Wsin(x)<Wx$. So it will be just like having a weaker spring when the pendulum makes a large angle, creating a longer period.