A simple pendulum with mass m on a string of length l is released from rest at an angle of $\theta_0$ to the vertical.

(i) Assuming that the pendulum undergoes simple harmonic motion, find an expression for its angular displacement as a function of time; hence determine its speed as it passes through the equilibrium position.

(ii) Using conservation of energy find this speed exactly and show that the result is the same as in (i).

So, I equated $-mg \sin \theta$ and $ml\ddot{\theta}$ for the first part and then set $\sin \theta = \theta$ using the small angle approximation. I then used $\theta = \theta_0 \cos(kt)$ as a general solution and worked out $k$ as $\sqrt{g/l}$. Then I got the velocity as $\sqrt{g/l}\theta_0$ with this method but I'm pretty sure that's wrong as my max velocity for part (ii) was $\sqrt{2gl(1-\cos \theta)}$ from the $E_k$ and $E_p$ equations (so I assume $\sqrt{2gl}$ as $\max E_p/E_k$.

Can someone tell me where I went wrong, thanks.


closed as off-topic by Aaron Stevens, user191954, John Rennie, rob Nov 6 '18 at 22:17

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  • $\begingroup$ I converted your expressions to MathJax (LaTeX code). Please use it next time in your questions, it makes them much more readable. $\endgroup$ – ahemmetter Nov 6 '18 at 15:25
  • $\begingroup$ Check the units on your velocity formula. That will tell you something. $\endgroup$ – Bill N Nov 6 '18 at 19:50