The problem is as follows:

The movement equation for a pendulum for some mass hanging in a weightless thread with the length $L$ is as follows:

$$\frac{d^2a}{dt^2} + \frac{g}{L}\sin(a) = 0$$

Where $a$ is the starting angle (in radians) and $g$ is the acceleration of gravity.

Determine the position of the pendulum after a second if $L = 0.2$m and if the starting angle of the pendulum is $3$ degrees and the velocity is $0$m/s.

You may also use the approximation $\sin(a) = a$.

My thoughts on it

Well, why not just move over the $(g/L)\sin(a)$ term to the right hand side and find the indefinite integral of both sides twice? I'm pretty sure that's not what you're supposed to do however and I'm frankly stumped. All the problems I've done so far has been of the form $f''(x) + f'(x) + f(x) = 0$.

The answer provided is: $a = (\pi/60)\cos\sqrt{5g}$ which is approximately $2.26$ degrees.

  • $\begingroup$ Use $\sin(a)\approx a$ and get a second-order linear differential equation. Find the solution using the contour conditions and give it as an approximate answer. $\endgroup$
    – Ian Mateus
    Dec 8, 2014 at 18:33
  • $\begingroup$ I think I mostly don't understand because g/L * a is a constant, at least as far as my understanding goes it is. I really need a complete solution to understand this because it feels like my mental model of this problem is entirely faulty and I need to know why. $\endgroup$
    – ejbs
    Dec 8, 2014 at 18:39
  • $\begingroup$ $a$ is a function of time. To compute the position, you need $a(1)$ and some basic geometry. $\endgroup$
    – Ian Mateus
    Dec 8, 2014 at 18:42
  • $\begingroup$ Yes but in the problem description they refer to a as both a function of time (d^2a/dt^2) and as a constant: "Where a is the starting angle (3 degrees)". Are you telling me that when they write sin(a) in the equation they actually mean sin(a(t)) and that a(0) = 3? $\endgroup$
    – ejbs
    Dec 8, 2014 at 18:47
  • $\begingroup$ Yes; it should be worded better, but that is precisely what is meant. $\endgroup$
    – BaronVT
    Dec 8, 2014 at 18:53

1 Answer 1


Using the small angle approximation i.e $\sin a \approx a$ we find $$ \ddot{a} + \sqrt{\frac{g}{L}}a = 0 $$ solutions to the above have the form $$ a(t) = A\sin\left(\sqrt{\frac{g}{L}}t\right) + B\cos\left(\sqrt{\frac{g}{L}}t\right) $$ and now we can place i.c into the solution $$ \dot{a} = A\sqrt{\frac{g}{L}}\cos\left(\sqrt{\frac{g}{L}}t\right) - B\sqrt{\frac{g}{L}}\sin\left(\sqrt{\frac{g}{L}}t\right) $$ at $t=0$ we find $$ \dot{a}(t=0)= A\sqrt{\frac{g}{L}} = 0 \implies A = 0 $$ thus the solution is of the form $$ a(t) = B\cos\left(\sqrt{\frac{g}{L}}t\right) $$ now since we have $a = 3^{\circ} = \frac{180^{\circ}}{60^{\circ}} = \frac{\pi}{60}$ we find $$ a(t=0) = B = \frac{\pi}{60} $$ therefore we have the solution $$ a(t) = \frac{\pi}{60}\cos\left(\sqrt{\frac{g}{L}}t\right) $$ now we need to find $a(t=1)$ plugging that in, and $L=0.2$, we find $$ a(t=1) = \frac{\pi}{60}\cos\left(\sqrt{\frac{g}{0.2}}\right) = \frac{\pi}{60}\cos\left(\sqrt{5g}\right) $$


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