Posted by how to do this? on Monday, January 30, 2012 at 1:11pm.
potential energy = m g h = m g L (1-cosA)
if angle A is small, cos A = 1 - A^2/2 +...
so
Pe = m g L A^2/2
kinetic energy = (1/2) m v^2 = (1/2) m L^2 [dA/dt]^2)
if A = sin wt
dA/dt = -w cos wt
max potential energy at top = max Ke at bottom
(1/2) m L^2 w^2 = (1/2) m g L
w^2 = g/L which by the way is in your book
g = L w^2
but w = 2 pi /T since when t = T, the angle w t must be 2 pi, full circle
g = L (2 pi)^2 / T^2
here L = .6 and T = 1.55
so g = 9.85 m/s^2
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