A vertical spring with mass is at equilibrium. The mass is then pushed up so that the spring is unstretched. The mass is dropped from this distance d from the equilibrium point. prove that the maximum distance that the spring stretches is 2d.

Question

anonymous · Answer

Is it definiely 2d and not $$\sqrt{2} d$$?
 
If you use GPE as $$mgd$$ then the systems total energy can be written as $$mgd = \frac{ 1 }{ 2 }mv ^{2} + \frac{ 1 }{ 2 }kx ^{2}$$. When the spring was stretched, at equilibrium position, the Force, mg caused the spring to stretch to d. So $$F = mg = kd$$ (I.e Hookes law). Now, to get the maximum x, the velocity will be zero (spring fully stretched out). This means the KE is zero in the big equation, plus use Hooke's law to substitute out the k, then solve for x. I get  sqrt(2) x d...