Quick Lagrangian of a pendulum question
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Mar5-12, 09:45 PM
I've got the first bit:
d2θ/dt2+(g/l)θ=0 in the small angle approximation,
which is S.H.M. with ω^2=√(g/l) (though I'm not sure about this as there's no minus sign in the E.O.M.) so T=√(l/g)
Careful here, [itex]\omega[/itex] is that angular velocity in radians per time unit. Your T is the time to travel one radian (of the oscillatory cycle) not time per cycle. You need to multiply by [itex]2 \pi[/itex].