1. Let $t_{s}$ be a scale for time. Write $\tilde{t}= \frac{t}{t_{s}}$ and consider $\tilde{\theta}(\tilde{t}) = \theta(t)$, not that the angle is noting dimensional already. Then using chain rule, $\frac{d^{2}}{d \tilde{t}^{2}} \tilde{\theta}(\tilde{t})= \frac{d^{2}}{d \tilde{t}^{2}} \theta(\tilde{t} t_{s}) = t_{s}^{2} \frac{d^{2}}{dt^{2}} \theta(t)$Therefore $0= \frac{d^{2}}{d t^{2}} \theta (t) + \frac{g}{l} \sin(\theta(t)) = \frac{1}{t_{s}^{2}} \frac{d^{2}}{d \tilde{t}^{2}} \tilde{\theta}(\tilde{t}) + \frac{g}{l} \sin (\theta(t))$so $0= \frac{d^{2}}{dt^{2}} \tilde{\theta}(\tilde{t}) + g \frac{t_{s}^{2}}{l} \sin (\tilde{\theta}(\tilde{t})). $Note $[l]=L$ and $[g]=LT^{-2}$ hence $[\sqrt{ l/g }]=T$. Therefore choose $t_{s} = \sqrt{ l/g }$ and obtain a DE w/out any parameters: $0=\frac{d^{2}}{d\tilde{t}^{2}} \tilde{\theta}(\tilde{t}) + \sin(\theta (t)).$