Ive been doing some integration study and ive been caught by this question. Anyone have any ideas? Thanks. Apologies on how the question is presented, im no quite sure how to do it properly yet.


It is known that for an ideal pendulum and for small initial displacement angle θ0 (from the vertical), the displacement angle $θ(t)$ at time t seconds is described by the differential equation $\frac{d^2θ}{dt^2}+ g /l θ(t) = 0$, $θ(0) = \theta_0$,$ \frac{dθ}{ dt} (0) = 0$, where $g$ is acceleration due to gravity and $ l$ is the length of the pendulum. By considering the Cosine function or otherwise solve the above differential equation. When will the pendulum first return to the vertical?


I'll write your ODE like $$\theta''(t) + gl \cdot \theta(t) = 0$$

Use the approach from the hint $$\theta(t) = A\cos(\omega t + \varphi)$$

which clearly solves the differential equation $$-\omega^2 A\cos(\omega t + \varphi) + gl \cdot A\cos(\omega t + \varphi) = 0$$

when $$\omega = \sqrt{gl} \ .$$

Now use $\theta(0) = \theta_0$ to compute the phase shift $$\theta(0) = A\cos(\omega 0 + \varphi) = A\cos(\varphi) = \theta_0 \ .$$

Use $\theta'(0) = 0$ to get $$\theta'(0) = -\omega A\sin(\omega 0 + \varphi) = -\omega A\sin(\varphi) = 0$$ so $$\varphi = 0 \ \ \Longrightarrow \ \ \theta_0 = A\cos(\varphi) = A\cos(0) = A \ .$$

Your final solution is $$\theta(t) = \theta_0 \cos(\omega t)$$ with $\omega = \sqrt{gl}$. This has improper units by the way, are you sure it's not $g/l$ on your sheet?

  • $\begingroup$ Lovely bit of maths, thanks very much. You correct, ill amend that now. $\endgroup$ – CianStanton Nov 26 '14 at 18:02

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