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Question Number 114340 by Dwaipayan Shikari last updated on 18/Sep/20

$$\frac{d^2\theta }{{dt}^2}+\frac{g}{l}sin\theta =0$$$$Or\stackrel{\cdot \cdot }{\theta }\left(t\right)+\frac{g}{l}\theta \left(t\right)=0$$

Commented by mr W last updated on 18/Sep/20

Commented by mohammad17 last updated on 18/Sep/20

$$whatsthemean\left(\frac{g}{l}\right)isitconstant?$$

Commented by Dwaipayan Shikari last updated on 18/Sep/20

Commented by Dwaipayan Shikari last updated on 18/Sep/20

$$Gravitationalaccelarationonsimplependulam(-g)$$$$Onhorizontaldirection-gsin\theta$$$$S=l\theta$$$$\frac{d^{2}s}{{dt}^2}=l\frac{d^2\theta }{{dt}^2}$$$$l\frac{d^2\theta }{{dt}^2}=-gsin\theta$$$$\frac{d^2\theta }{{dt}^2}+\frac{g}{l}sin\theta =0$$$$$$

Answered by Rio Michael last updated on 18/Sep/20

$$\mathrm{yes}!\stackrel{..}{\theta }\left(t\right)+\frac{\mathrm{g}}{l}\theta \left(t\right)=0$$$$\mathrm{an}\mathrm{experimental}\mathrm{solution}\mathrm{is}$$$$\theta \left(t\right)={\theta }_{\max }\cos (\omega t+\phi )$$

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