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

(d^2 θ/dt^2 )+(g/l)sinθ=0  Or     θ^(∙∙) (t)+(g/l)θ(t)=0

Commented bymr W last updated on 18/Sep/20

Commented bymohammad17 last updated on 18/Sep/20

whats the mean ((g/l))is it constant ?

Commented byDwaipayan Shikari last updated on 18/Sep/20

Commented byDwaipayan Shikari last updated on 18/Sep/20

Gravitational accelaration on simple pendulam  (−g)  On horizontal direction  −gsinθ  S=lθ  (d^2 s/dt^2 )=l(d^2 θ/dt^2 )  l(d^2 θ/dt^2 )=−gsinθ  (d^2 θ/dt^2 )+(g/l)sinθ=0

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

yes! θ^(..) (t) + (g/l) θ(t) = 0  an experimental solution is  θ (t) = θ_(max) cos (ωt + ϕ)