Question Number 123858 by Dwaipayan Shikari last updated on 28/Nov/20
$$\overset{..} {\theta}+\frac{{g}}{{l}}{sin}\theta=\mathrm{0}\:\:\:\:\:\left({Find}\:{the}\:{exact}\:{solution}\right) \\ $$
Answered by Olaf last updated on 29/Nov/20
$$\overset{\bullet\bullet} {\theta}+\frac{{g}}{{l}}\mathrm{sin}\theta\:=\:\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{1}\right)×\overset{\bullet} {\theta}\::\:\overset{\bullet\bullet} {\theta}\overset{\bullet} {\theta}+\frac{{g}}{{l}}\overset{\bullet} {\theta}\mathrm{sin}\theta\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\overset{\bullet\mathrm{2}} {\theta}−\frac{{g}}{{l}}\mathrm{cos}\theta\:=\:\mathrm{C}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{Usually}\:\mathrm{at}\:{t}=\:\mathrm{0},\:\theta\:=\:\theta_{\mathrm{0}} ,\:\overset{\bullet} {\theta}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{C}\:=\:−\frac{{g}}{{l}}\mathrm{cos}\theta_{\mathrm{0}} \\ $$$$\left(\mathrm{2}\right)\::\:\frac{\overset{\bullet} {\theta}}{\:\sqrt{\mathrm{cos}\theta−\mathrm{cos}\theta_{\mathrm{0}} }}\:=\:\sqrt{\frac{{l}}{{g}}} \\ $$$$…\mathrm{to}\:\mathrm{be}\:\mathrm{continued}… \\ $$