# Problem with a spring

1. Jan 4, 2012

### aaaa202

Look at the attached picture. I DO realize that this has been brought up several times, but there's a thing i still don't understand.
As you can see, I have solved the problem already but that was not without using several hints. Going through each hint at a time i solved the problem using conservation of energy after the collision. But but but! I don't agree with the energy assumptions they made. They made me use the elastical potentiel energy relative to the new equilibrium point as well as the kinetic energy and equaled that ½kA^2 where A denotes the amplitude.
BUT WHAT ABOUT THE LOSS IN POTENTIAL ENERGY DURING THE DOWNFALL? This depends on A as far as i can see making you need to solve a quadratic equation..
Anyways I got the right as you can see, but I don't get it...

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2. Jan 4, 2012

### SammyS

Staff Emeritus
So ...

What are you looking for from us?

3. Jan 5, 2012

### aaaa202

I said, that the solution did not involve using the fall in gravitational potential energy from the point where the ham and plate collide down to the lowest point of the oscillation. Why is that not used?

4. Jan 5, 2012

### SammyS

Staff Emeritus
It's not wrong to include the gravitational potential. However, all that including it does is to change the equilibrium position of the system. This merely represents adding a constant force.

The potential energy function, Vsg(y), with the spring and gravitational potentials included is:
$\displaystyle V_{\text{s g}}(y)=\frac{1}{2}ky^2+mgy$​
This function is quadratic in y, and has a minimum at $\displaystyle y=-\frac{mg}{k}\,.$ The coefficient of y2 is (1/2)k whether or not the gravitational potential is included.

This can also be seen by completing the square.
$\displaystyle V_{\text{s g}}(y)=\frac{1}{2}ky^2+mgy$

$\displaystyle =\frac{k}{2}\left(y*2+\frac{2mg}{k}y\right)$

$\displaystyle = \frac{k}{2}\left[y^2+\frac{2mg}{k}y+\left(\frac{mg}{k}\right)^2-\left(\frac{mg}{k}\right)^2 \right]$

$\displaystyle = \frac{k}{2}\left(y+\frac{mg}{k}\right)^2-\frac{m^2g^2}{2k}$
Compare this with the potential function, Vs(y), for a spring only.
$\displaystyle V_{\text{s}}(y)=\frac{k}{2}y^2$​

Adding a constant term to a potential function doesn't change the potential function's effect. So the only effect of adding the spring is to shift the equilibrium position.

5. Jan 5, 2012

### aaaa202

Hmm I still don't get it to be honest.
I do realize that the effect of gravity on a hanging spring is that it just changes the original equilibrium point of the spring - i.e. the motion is still SHM but around a lower point than had the spring laid horizontally.
But now you're saying as far as I can tell, that also the energy does nothing. So let's review: I have included the spring potential energy as well as the kinetic. What's the reason that the gravitational potential doesn't matter? Well okay you just explained that up there, but can you maybe do it less mathematically and more using intuitive, logical arguments. You've been a great help so far btw
Edit: thought about it a little more. If you include the gravitional potential the ham and plate would definately compress the spring more. But why on earth would they oscillate around another point? I mean, we did calculate a new equilibrium point when solving the assignment, which was using the equation mham + mplate = ky
So what equation would then describe this new equilibrium point, that you're talking about? :)

6. Jan 5, 2012

### SammyS

Staff Emeritus
No, it's not the case that the energy does nothing. Adding or subtracting a constant term to the potential energy function does nothing as far as the motion of a object is concerned. For instance, when using mgh for gravitational potential, it doesn't matter what point (elevation) you use for h=0 (the point at which gravitational P.E. is zero), what matters is the difference in P.E. from one position to another.

I'll try to get back with a couple of graphs of the potentials for this problem which may be helpful.

7. Jan 5, 2012