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Question Number 127974 by Dwaipayan Shikari last updated on 03/Jan/21

$$\stackrel{\bullet \bullet }{\theta }+\frac{g}{l}sin\theta =0$$$$Exactform\left(Mayincludeellipticintegral\right)$$

Commented by Dwaipayan Shikari last updated on 03/Jan/21

$$Mytry$$$$\stackrel{\bullet \bullet }{\theta }\stackrel{\bullet }{\theta }+\stackrel{\bullet }{\theta }\frac{g}{l}sin\theta =0\Rightarrow \frac{1}{2}.\frac{d}{dt}{\left(\stackrel{\bullet }{\theta }\right)}^2-\frac{g}{l}.\frac{d}{dt}\left(cos\theta \right)=0$$$$\Rightarrow {\left(\stackrel{\bullet }{\theta }\right)}^2-\frac{2g}{l}cos\left(\theta \right)=C$$$$\Rightarrow {\left(\stackrel{\bullet }{\theta }\right)}^2+\frac{4g}{l}{sin}^2\frac{\theta }{2}=C+\frac{2g}{l}\Rightarrow \stackrel{\bullet }{\theta }=\sqrt{C+\frac{2g}{l}-\frac{4g}{l}{sin}^2\frac{\theta }{2}}$$$$\Rightarrow \int \frac{d\theta }{\sqrt{C+\frac{2g}{l}}\left(\sqrt{1-\frac{4g}{C.l+2g}{sin}^2\frac{\theta }{2}}\right)}=\int dt$$$$\Rightarrow 2\int \frac{d\zeta }{\sqrt{C+\frac{2g}{l}}\left(\sqrt{1-{\left(\sqrt{\frac{4g}{C.l+2g}}\right)}^2{sin}^2\zeta }\right)}=t+\mathrm{\Phi }$$$$\Rightarrow 2\frac{1}{\sqrt{C+\frac{2g}{l}}}F(\frac{\theta }{2}\mid \sqrt{\frac{4g}{C.l+2g}})=t+\mathrm{\Phi }$$$$\Rightarrow F(\frac{\theta }{2}\mid \sqrt{\frac{4g}{C.l+2g}})=\frac{1}{2}(t+\mathrm{\Phi })\sqrt{C+\frac{2g}{l}}$$

Answered by mr W last updated on 03/Jan/21

$$\stackrel{\bullet }{\theta }\frac{d\stackrel{\bullet }{\theta }}{d\theta }=-\frac{g}{l}\sin \theta$$$${\left(\stackrel{\bullet }{\theta }\right)}^2=\frac{2g}{l}(\cos \theta -\cos {\theta }_0)$$$$\stackrel{\bullet }{\theta }=\frac{d\theta }{dt}=\sqrt{\frac{2g(\cos \theta -\cos {\theta }_0)}{l}}$$$$\frac{d\theta }{\sqrt{\cos \theta -\cos {\theta }_0}}=\sqrt{\frac{2g}{l}}dt$$$$\Rightarrow t=\sqrt{\frac{l}{2g}}{\int }_0^{\theta }\frac{d\theta }{\sqrt{\cos \theta -\cos {\theta }_0}}$$$$\Rightarrow t=\sqrt{\frac{2l}{g}}\times \frac{F(\frac{\theta }{2}\mid \frac{2}{1-\cos {\theta }_0})}{\sqrt{1-\cos {\theta }_0}}$$

Commented by Dwaipayan Shikari last updated on 03/Jan/21

$$Thankingyousir$$

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