The problem is as follows: Suppose the length of the pendulum is decreased by slowly pulling the string up through the hole. The rate of decrease of length is to be extremely slow compared with the frequency of oscillation, so that we can define a period of oscillation for each length of string. Find the change in energy of the pendulum as the length is shortened, from the work done in pulling against the tension in the string. Show as a result that the action variable J = E/ν is constant throughout the process. H = E = J/(2*PI) * (g/L)^1/2 From what I understand dE = J/(2*PI) * (g/(dL/dt))^1/2 and dv = 1/(2*PI) * (g/(dL/dt))^1/2 Hence dE= J dv J =dE/dv = E/v I'm not sure if this is correct. The question can't be this simple and I think I'm missing soming from the question. Also For a pendulum, if the amplitude of oscillation is small, show that the energy of the pendulum is given by E = Jν . I'm not sure how to get the right answer 0=theta 0' = theta dot T = m/2 * L^2 *0' ^2 V = -mgL cos 0 p = 0mL^2 H = p^2/2mL^2 -mgL cos 0 =E moving this around p = (2mL^2 (E +mgL cos 0)) ^1/2 subing cos 0 = 1 - (0 ^2) /2 then intergrating p I get J = 2*Pi * (m^2 *L^3 *g)^1/2 * (E/mgL +1/2) the answer would be correct if the +1/2 wasn't in the last term and I'm not sure where I hav maded a mistake in the intergration.