# Deriving a formula for max. speed of a simple pendulum bob

## Homework Statement

Derive a formula for the maximum speed V_max of a simple pendulum bob in terms of g, the length l and the maximum angle of swing Θ_max. (Hint: Use the fact that the same amount of energy at the top of the arc is the same as at the bottom of the arc.)

The picture depicts a string tied from the floor to the ceiling, and then hanging down from the ceiling most of the way down, which is swinging. It starts at the left hand amplitude. The distance from the left hand amplitude to the floor is labeled h = l-lcosΘ ; the distance to the ceiling from that point is labeled as lcosΘ. The angle is Θ and l is the distance of the string from the ceiling to where it hangs.

## Homework Equations

Θ = Θ_max * cos(εt+Φ)
ε = √(k/m) = √(g/l)
g = 9.8 m/s^2
Φ = arctan{-v(0)/[ε*x(0)]}
E = 0.5mv^2
E = mgh

## The Attempt at a Solution

I know that the answer, according to the book, is going to be v_max = √{2gl[1-cos(Θ)]} but I'm not sure if I got there correctly. The top of the pendulum would have the same energy as the bottom, so I suppose the E = 0.5mv^2 would mean that velocity is the max velocity. Where the pendulum swings, E = mgh, since it's at the amplitude?

Maybe if I set E = mgh = E = 0.5mv^2 and solve for v_max that will help.
mgh = 0.5mv^2
2gh = v^2
v_max = √(2gh)
v_max = √[2g(l-lcosΘ)]
v_max = √[2gl - 2glcosΘ]
v_max = √[2gl(1-cosΘ)]

^^^Does that all seem correct? Thank you!