The discussion centers on deriving the equations of motion for a mass-spring system involving two masses, m1 and M, with a spring constant k. The participants explore the differential equation $$m*\frac{d^2x}{dt^2} = -kx$$ and its solutions, emphasizing the significance of initial conditions and the role of gravitational force. Key insights include the necessity of expressing the equations in terms of the unknowns introduced, and the realization that the amplitude must exceed a certain threshold for m1 to take off from the platform.
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No, as I wrote in post #4 it depends how you define the position x=0. If you define it as being the equilibrium position then that is mg/k below the relaxed spring position. Thus the force in the spring is -k(x-mg/k) = -kx+mg. The net force on the object is thus (-kx+mg)-mg = -kx.RiotRick said:Update if anyone runs into the same problem. I don't have a solution but the attempt here is wrong. Right attempt would be ##x''m = -kx + m*g## Which leads to an inhom. diff. equation.
With this I close this thread o7