When: Lifting one mass one height equals one unit of energy... Lifting one mass two heights equals two units of energy... Lifting two masses one height equals two units of energy... Lifting two masses two heights equals four units of energy... Then: Dropping two masses two heights equals four units of energy!
Assuming SHM with period T & phase frequency ω, I get the work done by the spring on the falling mass from t=0 to ½T (y=y_max to 0 & force equilibrium at y_max): U= ¼k(y_max)² (Tω + 1) = ½k(y_max)² (π + ½). This is as opposed to the work done from static force, with no falling or SHM involved: U* = ½kX², with X= |0 - y_max|=|y_max|. The energy does depend on time PERIOD but the unit of time is cancelled out by the Phase frequency: ω=2π/T. Here I assumed ω is constant btw. Alternatively, you could say that the energy depends on the "phase" of the oscillation. So, prediction is that the ratio of spring dynamic to static restore energies is: U*/U= π+½≈ 3.64 The static spring work is really the potential energy difference and only depends on a difference in position. The dynamic spring work is the change in kinetic energy of the spring as the mass falls. ͔
@@Ian.Gostling I'm Ok thanks - you too I trust. Not yet reached the selling stage - so many snags to overcome week after week - a bit like building experiments - they always sound simpler in our heads, then we find the snags and endless modifications - now I'm being distracted by a mate trying to learn to use a new 3D printer and somehow thinks I can help but I;m clueless - learning fast through. Bet one would com in handy for some experiments. Could make Goofy a new brain.
Hey, what do you think of my proposal? Dynamic spring versus static spring? If the spring is "static" but the spring length has changed then I think that is equivalent to lowering the mass at a constant velocity. If the spring is "dynamic", then the mass is falling under gravity but is countered by the spring as it falls. The maths seems to indicate that a dynamic spring has more potential energy than the static case. About 3.64 times more! I'm not sure if there is a way of distinguishing the two results by experiment though. How can you measure this extra energy? A flywheel or something?
Goofy thinks that 'force' is the same as 'energy', but in reality it takes one joule of energy to lift a bottle of water one step up, but it takes twenty joules of energy to bring that same bottle all the way up to the next floor. While the water doesn't miraculously changes its mass during the experiment. 1 kg × 9.81 m/s² × 0.1 m = 0.981 J 1 kg × 9.81 m/s² × 2 m = 19.62 J Therefore: *E = m × g × h* Energy equals weight times height, thus!
