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

Problem diagram

Referring to the last figure if the angle $\theta$ is small we have that $y<<x$, 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?

  • 4
    $\begingroup$ 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. $\endgroup$
    – lemon
    Apr 13, 2015 at 14:32
  • $\begingroup$ Very insightful, Is there a way to do it more formally? Maybe with Taylor approx or something ? @lemon $\endgroup$
    – Keith
    Apr 13, 2015 at 14:39
  • $\begingroup$ 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 $\theta$? $\endgroup$
    – lemon
    Apr 13, 2015 at 14:47

1 Answer 1


Well I would do this.

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.

  • $\begingroup$ where $O(\theta^2)$ means of the order of $\geq \theta^2$ $\endgroup$
    – SuperCiocia
    Apr 13, 2015 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.