kashiark
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T=2pi√(l/g)
Is it in radian*seconds?
Is it in radian*seconds?
The equation for the period of a simple pendulum is T = 2π√(l/g), where T represents the period, l is the length of the pendulum, and g is the acceleration due to gravity. The units of √(l/g) are seconds per radian, while √(g/l) yields radians per second. The presence of 2π in the equation relates to angular motion, as it represents radians per revolution. The derivation involves understanding the forces acting on the pendulum and the relationship between angular frequency (ω) and linear displacement (x).
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kashiark said:Ok, that makes sense, but why is there a 2pi in the equation? What was the argument made when someone said, "Hey, I think period equals 2 pi times the square root of length over the acceleration due to gravity!"
kashiark said:Wow... I'm kind of overwhelmed. I can follow it from s=lΘ to d²Θ/dt²+(g/l)sinΘ=0 ... From there I'm too ignorant about differential equations to follow it, but why does F=-mgsinΘ ?
kashiark said:Wow... I'm kind of overwhelmed. I can follow it from s=lΘ to d²Θ/dt²+(g/l)sinΘ=0 ... From there I'm too ignorant about differential equations to follow it, but why does F=-mgsinΘ ?
kashiark said:Forget the first part, I just got that from thinking about it while brushing my teeth, but I still don't understand why it should be a negative mgsinθ
Edit: Ok, I figured that out as well, but I'm still stuck with the ω. It wasn't in any equation one second, and then it was. What does it represent anyway?
kashiark said:What's the x stand for?