[SOLVED] Pendulum Speed and total ME 1. The problem statement, all variables and given/known data The length of a simple pendulum is 0.84 m and the mass of the particle (the "bob") at the end of the cable is 0.68 kg. The pendulum is pulled away from its equilibrium position by an angle of 7.8 ° and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing? 2. Relevant equations E = .5mv^2+.5Iw^2+mgh+.5kx^2 w = sqrt(g/L) L-Lcos@ = max height of pendulum bob I of pendulum = (1/3)ML^2 3. The attempt at a solution so w = 3.416 rad/s The total energy: .5(.68)(v^2) + .5(1/3 ML^2)(w^2) + mgh + .5kx^2 kx^2 is potential energy, which is 0 at the bottom of the swing .34v^2 + (1/6)(.68*.84^2)(3.416^2) + .68*-9.8*.00777 .34v^2 + .88137 v^2 = 2.5922 v = 1.61 m/s Plugging this in gives a total ME of 1.7627 J The v and total E are wrong, but I'm not sure how to remedy this. Ideas, please? Thanks!