How can I derive the equation for the angle of a simple pendulum as a function of time without calculus?

Question

I want to derive the equation $\theta=\theta_0 \sin{\left(\sqrt{\frac{g}{L}} t\right)}$ from force diagrams and circular motion without using calculus. I have managed to get pretty close but I can't figure out how to obtain $\theta_0$ in the equation. I have checked other questions such as this one or this one, but none of them provide a detailed development. I have managed to reach the formula for the period $T=2\pi\sqrt\frac{L}{g}$ with trigonometry only, does that get me any closer?

Answer 1

Well, there is this complete answer which doesn't approximate anything and you probably should consider wasting some time on this and on real calculus (actually this would save you a lot of time), there is a lot of material online.

Let's say you've got a pendulum inclined from a small angle $\theta _0$, so small that $\sin (\theta) \approx \theta \approx \tan (\theta)$ and $\cos (\theta_0) \approx 1 - \frac{\theta_0^2}2$.

it's tangential acceleration is given by:

$a = \frac{P_ \theta}{m} = \frac{mg \sin (\theta)}m = g \sin (\theta) \approx g \theta$

notice that the acceleration is proportional to the distance of equilibrium, just like in an harmonic movement equation:

$a = - \frac km x = -\omega x$

so the movement will be like an MHS an the general equation will be:

$a = -\frac g L x \iff \omega = \sqrt{\frac g L}$ (the negative sign is because it goes faster as $x$ gets smaller, $x$ is the length of the circular arc $L \theta$). There is pure calculus behind this.