In the subject accelerometer (see link) the answer given is a = kgx. I get a = 2kgx. I'm wondering if anyone cares to discuss this problem. The answerer used an energy argument that I'm not sure I followed, whereas I used a simpler force-balance method. F = force on bead by parabola and is normal to the tangent at the quiescent point (F is in 1st quadrant on the exercise picture) θ = angle F makes with the x axis. 0 < θ < π/2 Fcosθ = ma Fsinθ = mg So F = mg/sinθ a = Fcosθ/m = (g/sinθ)cosθ = g/tanθ But θ = π - π/2 - arc tan(dy/dx) = π/2 - arc tan(dy/dx) where dy/dx = 2kx Applying tan(c-d) = [tan(c) - tan(d)]/[1 + tan(c)tan(d)] I get tanθ = 1/2kx giving a = g/tanθ = 2gkx. http://www.feynmanlectures.info/ Select "Exercises" and then "Bead parabola accelerometer".