# I What equilibrium distance to use for E of a vertical spring

1. Nov 2, 2016

I just wanted to confirm whether the idea I have about vertical springs is correct or not.

Suppose you have an ideal (massless) spring oriented vertically, and suspended from a block in the air, which in turn is mounted on a stand and placed on a desk. The distance from the desk (or the base of the stand) to the equilibrium resting point of the floating end of the spring, with NO masses attached at the end of the spring, is y0. Now a mass is attached to the end of the spring, and it extends to a new equilibrium point called y1 (again measured from the desk).

Now suppose that we set this mass and spring system to oscillate by pulling downwards on it. The question is, at some arbitrary position y2 in the mass' motion, what is the equation for the total mechanical energy? Just for simplicity lets say y2 is the minimum distance; i.e. when the spring is maximally stretched.

My answer is the following - I just wanted to know whether this is correct.

[1] E = (1/2)mv2 + (1/2)k(y2 - y0)2 + mg(y2)

There are other sources online that indicate that it is alright to write the total energy in the following way, and that "nothing will change":

[2] E = (1/2)mv2 + (1/2)k(y2 - y1)2 + mg(y2)

To me it seems extremely clear that the total energy of the system is only correctly described by equation [1]. Equation [2] is ignoring the stored potential energy in the spring from y0 to y1.

Am I missing anything? Clearly both answers can not be correct. It seems utterly bizarre to me that equation [2] is correct since it is explicitly ignoring stored energy.

Moreover if we were to look at differences in energy at two points, designated y1 and y2 (so yes, one of the points is at the new equilibrium distance), in case [2], the change in energy will be proportional to:

[2] (y2 -y1)^2 - (y1-y1)^2 = (y2-y1)^2

While for case [1], the change in energy will be proportional to:

[1] (y2 - y0)^2 - (y1 - y0)^2

Assuming that, regardless of how one 'defines' the equilibrium point, the velocities calculated at y2 and y1 are the same in either case, and that the difference in gravitational potential energy is the same, it seems clear to me that case [1] yields a vastly different answer from [2] due to cross-terms involving y0.

So given that there is clearly a different change in energy depending on how one defines the equilibrium point for a vertical spring, it seems that only one case can be correct. Again I ask, is case [1] correct as I believe it to be?

This should imply that it is therefore totally incorrect to use y1 as a "new equilibrium" for calculating total mechanical energy.

Thank you for the assistance.

Edit: I just did a computer simulation in Python with the following code (Python version 2.8 or so):

C = 0

y0 = 10 + C
y1 = 9 + C
y2 = 8 + C

def new():
return (float(y2-y1))**2

def old():
return (float(y2-y0))**2 - (float(y1-y0))**2

print "New", new()
print "Old", old()

For New I got 1 and for Old I got 3 (I used floats just in case something funky happened with integers; i.e. the division issue for integers..). The value of C doesn't matter as it shouldn't. So there is clearly a difference. Only one can be correct. I guess that answers it :). But if somehow I have done something else wrong I would like to hear about it from the community.

Last edited: Nov 2, 2016
2. Nov 2, 2016

### olivermsun

When you dangle the mass from the spring, the system droops from y0 to a new equilibrium point y1, thereby increasing the energy stored in the spring.

So what happens to the potential energy of the mass relative to the point from which you released it?

3. Nov 3, 2016

The potential energy of the mass will decrease. In principle, in order to satisfy the conservation of energy, the energy which is increased when it gets stored in the spring is cancelled out by the gravitational potential energy which is lost, assuming you are at equilibrium at both times (so speed is zero and kinetic energy plays no role).

But I guess the point remains, that the gravitational potential energy which is lost has to be accounted for *fully* by the gain in the spring's potential energy. If you consider the new equilibrium point of the spring to be y1, then its "new" potential energy is just 1/2k(y1-y1)2 = 0, which is impossible as it implies no energy was lost through Eg = mgh

I think you are pointing me in the direction that says, the equilibrium position, from which the potential energy of any spring is measured, MUST be y0. That certainly helps clarify things :). I guess I just find it hard to trust my own intuitions. But this is so overwhelmingly clear that I feel I have resolved the problem.

Hey that was a really easy way to point out the issue. I didn't need to do all that code at all. Thanks

Last edited: Nov 3, 2016
4. Nov 4, 2016

### vanhees71

You can solve for it by symmetry. For a homogeneously charged plane in the $xy$ plane of a Cartesian coordinate system, you expect the electric field to be of the form
$$\vec{E}=E_0 \text{sign} z \vec{e}_z$$
due to symmetry. Now choose a cube parallel to your cartesian coordinates enclosing a part $a^2$ of the xy-plane. Then you get
$$Q=2 E_0 a^2=\sigma a^2 \; \Rightarrow \; E_0=\frac{\sigma}{2}.$$
The potential of the electron in this electric field is
$$V=e \vec{x} \cdot \vec{E}=e \frac{\sigma |z|}{2}, \quad e>0.$$
Suppose it's going with velocity $v_z$ in positive $z$ direction (starting at a point $z=z_0>0$) then it's easy to get how far it comes before falling back by using energy conservation.
$$\frac{m}{2} v^2+V(z_0)=V(z_{\text{max}}) \; \Rightarrow \; e \frac{\sigma}{2} (z_{\text{max}}-z_{0})=\frac{m}{2} v^2 \; \Rightarrow \; z_{\text{max}}=z_{0}+\frac{m v^2}{e \sigma}.$$

5. Nov 7, 2016

### nasu

How is this related to the spring question?

6. Nov 8, 2016

### vanhees71

Argh. I don't know how this happened, but this is the answer to another question in another thread. Sorry for the confusion.