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If I attempt to find the displacement $x$, from equilibrium when a mass, $M$ is attached to an upside down vertical spring with spring constant $k$ I get two different answers when I use energy considerations and when I use forces. There is probably some simple conceptual mistake I am making but I can't identify it. Here are the equations I am referring to

$0=\frac{1}{2}kx^2-mgx$, (making our equilibrium point, $y = 0$) $$x=\frac{2mg}{k}$$ Now using forces, $$mg=kx$$ $$mg/k=x$$ Obviously $2mg/k$ is not equal to $mg/k$ so what is my mistake?

Qmechanic
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LeBronJames
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1 Answers1

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What you get from the first equation is the maximum displacement of the mass. When you drop the mass, that is the lowest it will go and then it will continue oscillating (in absence of any damping). The second equation gives you the position of static equilibrium.

Here's one (hand-wavy) way to see these things:

There is no non conservative force acting on the system and so the total energy, $T$, is conserved. $$T=KE_{mass}+PE_{spring}+PE_{mass}$$ As the mass moves downward, it loses potential energy and gains kinetic energy as well as transfers some potential energy to the spring. If you choose to apply the energy conservation at the static equilibrium position, you have to take into account the velocity at that point as well since it will zero. However, at the point of maximum displacement, the mass stops momentarily and then you can say that the drop in gravitational potential energy is equal to the increase in potential energy of the spring.

ModCon
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