# To solve the equations of motion of a simple Pendulum $\ddot \theta = -\dfrac gl \sin \theta$

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Hello,

I've a problem to calculate the Position of a pendulum as a function of theta.
For example: $\theta (t)$ is a function of time which returns the angle made by the pendulum at a particular instant wrt it's equilibrium Position.

So,
$$T = \dfrac 12 m l^2 \dot \theta^2$$$$U = - mgl \cos \theta$$
$$L(\theta, \dot \theta) = \dfrac 12 m l^2 \dot \theta^2 + mgl \cos \theta$$
Using, the Euler - Lagrangian Formula,
$$\dfrac d{dt} \left ( \dfrac{\partial L}{\partial \dot \theta}\right) - \dfrac{\partial L}{\partial \theta} = 0$$
We get,

$$\boxed{\ddot \theta =- \dfrac gl \sin \theta}$$
which is the equation of motion.
But, most of the derivations, I've seen/read go this way:
$$\ddot \theta = \dfrac gl \theta \quad \dots \quad (\text{As, } \sin \theta \approx \theta, \theta \rightarrow 0) \tag{*}$$

$$\theta (t) = \cos \left ( \sqrt{\dfrac gl} t \right)$$
Because it satisfies $(*)$

So, I've 2 questions here.

1. Other possible solutions of the Second Order Differential Equations exist like $\theta (t) = e^{\left( \sqrt{\dfrac gl}t \right)}$. So,
why we choose that only one? One would argue that the sine function
oscillates similar to the pendulum, so this makes sense to accept the
sine one. But, in general case, when we solve the Lagrangian and get
the equation of motion in differential form, then there are tons of
complex situation possible, How can you determine which kind is
needed?

2. How can we solve the Second Order Differential Equation $\ddot \theta = - \dfrac gl \sin \theta$ and get an exact formula for that?

Thanks :)

asked May 11
The cosine solution satisfies (*) with a minus sign on the RHS. Whether you choose sine or cosine, or a particular linear combination of them, depends on the initial condition.

## 1 Answer

1 vote

You left out a minus sign in the 2nd equation. It should be written $$\ddot \theta \approx -\frac{g}{l}\theta$$ Now you can see that $\theta=e^{\sqrt{\frac{g}{l}} t}$ is not a solution because $\ddot \theta =+\frac{g}{l}\theta$. Instead we need a solution of the form $\theta=\pm e^{i\sqrt{\frac{g}{l}}t}$. Then we get $$\ddot \theta = i^2\frac{g}{l}e^{i\sqrt{\frac{g}{l}}t}=-\frac{g}{l}\theta$$

Solutions to $$\ddot \theta = \pm\frac{g}{l}\theta$$ have the form $\theta=e^{\pm \sqrt{\frac{g}{l}} t}$ when the sign on the right is +ve and $\theta=e^{\pm i\sqrt{\frac{g}{l}} t}$ when the sign on the right is -ve. The most general solution is a linear combination : $$\theta_{+}=Ae^{+\sqrt{\frac{g}{l}} t}+ Be^{-\sqrt{\frac{g}{l}}t}=C\cosh(\sqrt{\frac{g}{l}} t+\phi)$$ $$\theta_{-}=Ae^{i\sqrt{\frac{g}{l}}t}+Be^{-i\sqrt{\frac{g}{l}} t}=C\cos(\sqrt{\frac{g}{l}} t+\phi)$$ for suitable values of $C, \phi$.

Your 2nd equation $$\ddot \theta \approx -\frac{g}{l}\theta$$ is linear and is called the Harmonic Equation. Solutions are simply linear combinations of exponentials or trigonometric functions.

Your 1st equation $$\ddot \theta \approx -\frac{g}{l}\sin\theta$$ is non-linear. Solutions cannot be expressed in terms of simple exponential and trigonometric functions. They can be expressed in terms of elliptic functions as shown in https://www.math24.net/nonlinear-pendulum/.

answered May 14 by (28,806 points)
edited May 15