Question Number 201822 by MathedUp last updated on 13/Dec/23
$$\mathrm{Do}\:\mathrm{Not}\:\mathrm{Use}\:\mathrm{sin}\left(\theta\right)\sim\theta\:\left(\theta\:\:\mathrm{is}\:\mathrm{small}\:\mathrm{Enough}\right) \\ $$$$\ddot {\theta}+\frac{\mathrm{g}}{\ell}\mathrm{sin}\left(\theta\right)=\mathrm{0} \\ $$$${y}''\left({t}\right)+\frac{\mathrm{g}}{\ell}\:\mathrm{sin}\left({y}\left({t}\right)\right)=\mathrm{0} \\ $$$${y}''\left({t}\right){y}'\left({t}\right)+\frac{\mathrm{g}}{\ell}\mathrm{sin}\left({y}\left({t}\right)\right){y}'\left({t}\right)=\mathrm{0} \\ $$$${y}'\left({t}\right){y}''\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left({y}'\left({t}\right)\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{g}}{\ell}\mathrm{sin}\left({y}\left({t}\right)\right){y}'\left({t}\right)=−\frac{\mathrm{g}}{\ell}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\mathrm{cos}\left({y}\left({t}\right)\right) \\ $$$$\therefore\:\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left({y}'\left({t}\right)\right)^{\mathrm{2}} −\frac{\mathrm{g}}{\ell}\mathrm{cos}\left({y}\left({t}\right)\right)\right]=\mathrm{0} \\ $$$$\therefore\frac{\mathrm{1}}{\mathrm{2}}\left({y}'\left({t}\right)\right)^{\mathrm{2}} −\frac{\mathrm{g}}{\ell}\mathrm{cos}\left({y}\left({t}\right)\right)={c}_{\mathrm{1}} \:\:\boldsymbol{\mathrm{Const}} \\ $$$${y}''\left({t}\right)+\frac{\mathrm{g}}{\ell}\mathrm{sin}\left({y}\left({t}\right)\right)=\mathrm{0}\rightarrow\left({y}'\left({t}\right)\right)^{\mathrm{2}} −\frac{\mathrm{2g}}{\ell}\mathrm{cos}\left({y}\left({t}\right)\right)={c}_{\mathrm{1}} \\ $$$$\mathrm{and}…\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{Sove}\:\mathrm{Diff}\:\:\mathrm{Equa}.. \\ $$
Commented by mr W last updated on 13/Dec/23
Commented by mr W last updated on 13/Dec/23
Commented by mr W last updated on 14/Dec/23
$${not}\:{only}\:{you},\:{but}\:{all}\:{people}\:{in}\:{the} \\ $$$${world}\:{can}\:{not}\:{solve}\:{it}\:{using}\:{elementary} \\ $$$${functions}.\:{to}\:{find}\:{the}\:{period}\:\:{you}\: \\ $$$${can}\:{apply}\:{elliptic}\:{integrals}. \\ $$
Commented by MathedUp last updated on 14/Dec/23
$$ \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{a}\:\mathrm{lot} \\ $$