1. The problem statement, all variables and given/known data A Particle of mass m is threaded on a frictionless rod that rotates at a fixed angular frequency Ω about a vertical axis. A spring with rest length Xo and spring constant k has one of it's ends attached to the mass and the other to the axis of rotation. Let x be the length of the spring in it's stretched condition and thus the distance of the mass from the axis of rotation. Assume k is greater than mΩ^2 a. Construct a Lagrangian for this system b. Find the equation of motion of the mass c. Find Xeq, the equilibrium position of the mass d. Find the frequency of small oscillations about this equilibrium point 2. Relevant equations L= T - U ∂L/∂x = d/dt (∂L/∂x') 3. The attempt at a solution I have my solution attached as a copy. I am not trying to find a solution, but to understand what the solution means, and how to make sense of it. It is part c and part d that I am having trouble understanding.. A. L = 1/2m(x'^2 + x^2Ω^2) - 1/2k(x - Xo) B. mΩ^2x - k(x-Xo) = mx'' (x' = dx/dt, ect..) C. Xeq = (-kXo)/(mΩ^2-k) (because x' = x'' = 0) D. Letting x = Xeq + ε → x'' = ε'' Then he writes here, mε'' +(k-mΩ^2)ε = 0 Which has the form of the simple harmonic oscillator, mx'' +kx = 0 So the frequency is just like in the simple harmonic oscillator, ω^2 = (k-mΩ^2)/m My only questions concern the last two steps. What is going on for part C I do understand that if the mass is in equilibrium than it does not have an acceleration or a velocity, so that follows naturally. I do not understand part D, because in the equation of motion, kXo has vanished, which I do not understand. x = Xeq + ε → x'' = ε'' I don't understand how this here implies that I can just ignore that Xo term. I'm just not really sure what's going on with this last part here. If anybody could explain this to me it would be of great use to me. Thank you.