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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