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A simple pendulum

A simple pendulum

The equations of motion of a simple pendulum can easily be derived using Newtonian physics.

Let $\theta$ be the angular displacement from the central, equilibrium position. Let $u$ be the velocity of the bob when $\theta = 0$ and let the bob have a mass of $m$, with the support being of negligible mass and of length $l$.

By applying $F=ma$ to the unpaired tangential force, we get $-mg\sin\theta = ma$, (note the sign is due to the force being in the opposite direction to that generated by increasing $\theta$) we can now determine a second expression for $a$.

Recall that $a = \ddot{x}$, where $x$ is displacement. In this case, $x = l\theta$, differentiating this gives $\dot{x} = l\dot{\theta}$, differentiating a second time yields $\ddot{x} = l\ddot{\theta}$.

Substitution of this yields the differential equation

$$-\frac{g}{l}\sin\theta = \frac{d^2\theta}{dt^2}$$

It is worth noting that this equation doesn't have an analytical solution. Instead the approximation $\sin(\theta) \approx \theta$ (for small values of $\theta$) gives a second order linear ordinary differential equation,

$$\frac{d^2\theta}{dt^2} + \frac{g}{l}\theta = 0$$

which is then analytically solvable, with solution

$$\theta = A\cos\left(\sqrt{\frac{g}{l}}t\right) + B\sin\left(\sqrt{\frac{g}{l}}t\right)$$.

The values of $A$ and $B$ depend on the initial values of the particular system under study.

For example, consider the pendulum system with $l=1m$, $M = \frac{1}{10}kg$, $\theta_0 = \frac{\pi}{12}$ and $\dot{\theta}_0=0$ and $g = 9.8ms^{-2}$. Then, assuming we can linearise the underlying differential equation and use the general solution above,  we get

$$\theta = A\cos\left(\sqrt{\frac{9.8}{1}}t\right) + B\sin\left(\sqrt{\frac{9.8}{1}}t\right)$$

Differentiating this equation gives

$$\dot{\theta} = -\sqrt{9.8}A\sin\left(\sqrt{\frac{9.8}{1}}t\right) +\sqrt{9.8}B\cos\left(\sqrt{\frac{9.8}{1}}t\right)$$

Applying $\theta_0 = \frac{\pi}{12}$ and $\dot{\theta_0} = 0$ gives

$$\frac{\pi}{12} = A\cos\left(\sqrt{\frac{9.8}{1}}\times 0\right) + B\sin\left(\sqrt{\frac{9.8}{1}}\times 0\right) = A$$

$$0 = \sqrt{9.8}B\cos\left(\sqrt{\frac{9.8}{1}}\times 0\right) = \sqrt{9.8}B$$

The motion is therefore described by

$$\theta = \frac{\pi}{12}\cos\left(\sqrt{9.8}t\right)$$