0
$\begingroup$

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?

enter image description here

$\endgroup$
6
  • 1
    $\begingroup$ It would help to know exactly where you are applying the small angle approximation. I would guess to Newton's laws of motion, right? $\endgroup$
    – user196418
    Dec 17, 2018 at 12:58
  • 1
    $\begingroup$ Where is your origin? That matters a lot too. $\endgroup$
    – QuIcKmAtHs
    Dec 17, 2018 at 13:34
  • 1
    $\begingroup$ $sin(\theta/2)=\theta/2-1/3!((\theta)/2)^3+....$ so for small angles like 6 1/3!((\theta)/2)^3=1.9*10^-4 $\endgroup$ Dec 17, 2018 at 13:38
  • 1
    $\begingroup$ Is this the original diagram that was used? Maybe, in the original diagram, theta, y, and x were defined differently? $\endgroup$ Dec 17, 2018 at 13:49
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 17, 2018 at 13:50

1 Answer 1

2
$\begingroup$

It seems like the origin is not P, but the point where the pendulum intersects the vertical axis. As a result the coordinates are:

$x = L \sin(\theta)$

$y = L - L \cos(\theta)$

Which gives us:

$\frac{y}{x} = \frac{1 - \cos(\theta)}{\sin(\theta)} = \tan(\frac{\theta}{2}) \approx \frac{\theta}{2}$

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 $\frac{\theta}{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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