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

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).

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.
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.