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1

To shamelessly steal what James says above: the scale doesn't measure your mass, which remains the same no matter where you are, or what movements you make. The scale measures your weight, which is your mass multiplied by the acceleration due to the Earth's gravity, acting between your feet and the base of the scale. You will measure your correct weight ...

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The obvious solution is to add a "dashpot damper" to the mix so that the equation of motion of the surface is $$y''(t) + 2 \lambda ~ y'(t) + \omega^2 y(t) = -\alpha w_0.$$As usual, in this case the equilibrium comes to a height $y_0 = -\alpha w_0 / \omega^2,$ which reads on the scale as the weight $w_0.$ Substituting $y = y_0 + \eta$ gives a function ...

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There is the instantaneous force between the mass and the scales, and then there's the reading of the scales. Factors influencing both of these depend on many unknown factors - so here are just some general thoughts. First, if you drop a mass $M$ from height $h$ onto scales with mass $m$, and the two will then move as one, the collision is considered ...

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