# Simple pendulum cartesian coordinates

I'm reading Vibrations and Waves from French. I don't understand the following approximation when considering the simple pendulum:

Referring to the last figure if the angle $$\theta$$ is small we have that $$y<, from the geometry of the figure:

$$y\approx \frac{x^2}{2l}$$

where $$l$$ is the length of the string.

I understand this, the following is what I don't get:

The statement of the conservation of the energy is :

$$E= \frac{1}{2}mv^2+mgy$$

where $$v^2=\Big(\frac{dx}{dt}\Big)^2 +\Big(\frac{dy}{dt}\Big)^2$$

given the approximations already introduce is very nearly correct to write:

$$E= \frac{1}{2}m\Big(\frac{dx}{dt}\Big)^2+mg\frac{x^2}{2l}$$

My question is:

why do they neglect the $$y$$ component of the velocity ?

Can somebody explain to me what approximations do they use?

• It follows geometrically. When $\theta=0$, the velocity will be entirely horizontal (i.e. $v_y=0$). And when $\theta$ is close to $0$, $v_y$ will also be close to zero. Apr 13, 2015 at 14:32
• Very insightful, Is there a way to do it more formally? Maybe with Taylor approx or something ? @lemon Apr 13, 2015 at 14:39
• Well you can see that $v_y$ will be proportional to $\sin(\theta)$, so can you formalise the behaviour of $\sin(\theta)$ in the limit of small $\theta$? Apr 13, 2015 at 14:47

The velocity vector $\mathbf{v}$ is $$\mathbf{v} = (v_x, v_y) = (v\cos\theta, v\sin\theta),$$ where $v = |\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$.
Small angle approximation means that $\cos\theta \approx 1 - \frac{\theta^2}{2} + O(\theta^4)$ and $\sin\theta \approx \theta + O(\theta^3)$.
But you are interested in the square of the velocity (you want the kinetic energy), and you notice that $v_x^2 \propto 1 + O(\theta^2)$ and $v_y^2 \propto O(\theta^2)$, and to first order you only keep the terms that go in $\theta^1$ so effectively you drop the $v_y$ component.
• where $O(\theta^2)$ means of the order of $\geq \theta^2$ Apr 13, 2015 at 15:00