[Physics] Hanging mass from spring/ Setting potential to 0

forceshomework-and-exercisesnewtonian-mechanicspotential energyspring

I'm working a problem from Taylor's Classical Mechanics Book, and it's highlighted a couple issues I never quite wrapped my head around (despite getting good marks in advanced undergrad mechanics and EM).

Here are the questions:

a) Show that a spring obeying Hooke's Law has a corresponding potential energy $$U = \frac{1}{2}kx^2$$ if we choose $$U$$ to be zero at the equilibrium position.

b) If this spring is hung vertically with a mass $$m$$ suspended from the other end and constrained to move only in the vertical direction, find the extension $$x_0$$ of the new equilibrium position. Show that the total potential (spring plus gravity) has the same form $$\frac{1}{2}ky^2$$ if we use the coordinate $$y$$ equal to the displacement measured from the new equilibrium position at $$x=x_0$$, and redefine our reference point so that $$U=0$$ at $$y=0$$.

What I Know

I know that we define the potential of a field as the work done to move an object through the field from a reference point $$x_0$$ at which we define the potential to be $$0$$

$$U(\vec{r}) = -\int_{\vec{r_0}} ^{\vec{r}}\vec{F}\cdot{d\vec{r}}~~$$

What I am not clear on is what we are actually doing mathematically when we "set $$U =0$$". I've thought of it several different ways and I've never been super clear on which is correct:

1) I've thought about it as an application of the fundamental theorem of calculus
$$U(\vec{r}) = -\int_{\vec{r_0}} ^{\vec{r}}\vec{F}\cdot{d\vec{r}} = -[U(\vec{r}) – U(\vec{r_0})]~~$$

This last equality is where I'm a little confused (obviously because I just proved that something equals its own additive inverse)

Now, $$U(r_0)$$ vanishes, because we chose it to be zero.

Edit it has been pointed out that this is faulty because the antiderivative of the force evaluated at a point is not by itself a potential.

2) Alternatively, we change our coordinate system such that we place the origin where $$U = 0$$. Assuming all potentials depend on position such that the potential is zero when $$\vec{r} = 0$$, this works nicely, but I don't know if there are exotic potentials that depend on $$\vec{r}$$ some other way.

My attempt at solutions

1)

If I fix my coordinate axes such that the non-fixed end of the spring lies at the origin ($$\vec{r}= 0$$), we get that the work done to stretch the spring is:

$$\int_0^x \vec{F} \cdot d\vec{r} = -\int_0^x kx'dx' = -\frac{kx'^2}{2}\big|_0 ^ x = -\frac{kx^2}{2}$$.

So that works, but I'm not clear if I'm correctly "setting $$U$$ to zero at the reference point.

2)

Finding the new equilibrium position is simple; I need to know when the force due to gravity is equal to the force of the spring; that is when

$$-kx = mg$$ (defining force positive downwards).

So our new $$x_0 = \frac{mg}{k}$$

The next part is quite confusing to me; if we set our new zero potential to the new equilibrium position, and then the force on the spring is $$mg – ky$$, where $$y$$ is the displacement from this new equilibrium and I'm eliding the fact that these are vectors since we're constrained to 1-D movement. Then the potential for any given displacement is $$\int_0^y (mg- ky )dy = mgy -\frac{ky^2}{2}$$. But I'm trying to show that the total potential is of the form $$-\frac{ky^2}{2}$$. What am I missing?

Edit it was pointed out that the force exerted by the spring is actually dependent on its displacement from its resting length. Thus the potential should be $$\int_0^y (mg- k(y-\frac{mg}{k})dy = -\frac{ky^2}{2}$$

Any clarification on these topics would be greatly appreciated.

For the spring without mass attached, the PE will be $U(x)= -\int_0^xF(x)dx = -\int_0^x kx dx = -\frac{1}{2} kx^2|^x_0=-[\frac{1}{2} kx^2-0]=-\frac{1}{2} kx^2$
This is just a special case of the more general: $\Delta U= -\int_{x_1}^{x_2}F(x)dx = -\frac{1}{2} kx^2|^{x_2}_{x_1}=-[\frac{1}{2} kx_2^2- \frac{1}{2} kx_1^2]$