Discrepancy in the spring constant using force vs energy

[MrRice5555]

Lets say you have a spring with unknown constant k. You try to calculate the value of this constant by hanging a known mass m from the spring. The spring stretches until the force of the spring equals the weight of the mass. Thus, kx = mg, and k = mg/x . However, if you try to calculate this same constant using energy instead, by setting the zero point at the lowest point after the spring is stretched, we have an initial potential energy of mgh. Once you release the mass though, the potential energy of gravity converts into potential energy of spring, which is 0.5 kx^2. Setting mgh = 0.5kx^2, we have k = 2mgh / x^2. However, we defined h = x, and the equation reduces to k = 2mg/ x, exactly twice what we calculated using force. I don't understand why this happens, because your answer should be the same no matter how you calculate it. If you can explain what mistake I'm making, that would be very helpful. Thanks!

--Stephen

 

[kcdodd]

The force is not zero when the total potential energy is zero. In short, x is not the same in your two methods.

Taking the total potential energy as U = 0.5*k*x^2 + mgx, with x=0 at the equilibrium of the spring by itself. Then the there are two roots of U = 0; x = 0, and x = -2*m*g/k. However, that does not mean the net force is zero. F = -dU/dx = -(kx + mg). At the roots of U=0 you get F = -mg and F = mg, not F=0. The position which F=0 is the position around which the mass will oscillate.

 

[Andy Resnick]

Lets say you have a spring with unknown constant k.

--Stephen

Heh... this is a great paradox! It's kept me up past my bedtime...

The conceptual error is in treating an equilibrium configuration as a process:

At equilibrium the sum of the forces (and torques) is zero; also, the energy is a minimum.

The sum of the forces: kx = mg, k = mg/x,
The minimization of the energy: 1/2kx^2 -mgx = U, dU/dx = 0 -> kx - mg = 0 -> k = mg/x.

Note the sign of mgx- the weight moves down from x = 0.

 

[Studiot]

Here is another explanation.

You are trying to equate the work done by the weight on the spring to the loss of potential energy of the weight.

This is not correct because it is missing a term.

Consider what happens when when you release the weight:

The weight falls gaining kinetic energy. Some of this is transferred to the spring but when the weight reaches the position where the spring return force just balances the gravity force on the weight, the weight is still moving.

So the true energy balance is

Potential Energy lost by weight = Elastic Energy gained by spring + Kinetic energy gained by weight.

This latter lerm leads to the oscillation about the equilibrium position.