# Small Angle Approximation for Simple Pendulum

I am working on a simple pendulum problem. The $$y$$ direction is vertical and the $$x$$ direction is horizontal. Displacement in the $$x$$ direction is taken to be much less than the length of the string, $$L$$.

One of the small angle approximations given for this problem was $${\theta \over 2} \approx {y \over x}.$$ where $$y$$ and $$x$$ represents the coordinates of the pendulum.

Why is this true? One of the small angle approximations I know is $$\tan \theta \approx \theta,$$ giving $$\frac{x}{y}\approx\theta.$$

Where did the factor of two in the first equation come from?

• It would help to know exactly where you are applying the small angle approximation. I would guess to Newton's laws of motion, right? – ggcg Dec 17 '18 at 12:58
• Where is your origin? That matters a lot too. – QuIcKmAtHs Dec 17 '18 at 13:34
• $sin(\theta/2)=\theta/2-1/3!((\theta)/2)^3+....$ so for small angles like 6 1/3!((\theta)/2)^3=1.9*10^-4 – Abdelrhman Fawzy Dec 17 '18 at 13:38
• Is this the original diagram that was used? Maybe, in the original diagram, theta, y, and x were defined differently? – Chet Miller Dec 17 '18 at 13:49
• If you look carefully the geometric scheme you should realize that, with the notation uset there, $tan \theta = y/x$. Of course, for small angles $tan \theta$ and $\theta$ can be identified. – GiorgioP Dec 17 '18 at 13:50

$$x = L \sin(\theta)$$
$$y = L - L \cos(\theta)$$
$$\frac{y}{x} = \frac{1 - \cos(\theta)}{\sin(\theta)} = \tan(\frac{\theta}{2}) \approx \frac{\theta}{2}$$
Another way of finding the same result is to calculate (geometrically) the angle between the pendulum, the origin, and the horizontal axis: it is equal to $$\frac{\theta}{2}$$.