# Homework Help: Finding Spring Constant from a slope with a constant x?

1. Nov 13, 2012

### cameronjrhea

1. The problem statement, all variables and given/known data
I quickly copied this down, so it is paraphrased but I hope it still makes sense.

Some children are playing with a spring toy, compressing it off the ground and seeing how high it bounces. When they attach different masses to it, it bounces different heights. Each time they compress the spring to exactly .02 meters. What would the children would have to graph where as the slope of the line is k, the spring constant. the mass of the spring is negligible.

Basically, if you have a constant x, but a varying height and mass, how do you find the spring constant? and what would you graph to make that the slope? I am having lots of trouble with this

2. Relevant equations
F=kx
1/2kx^2 = mgh

I really don't know

3. The attempt at a solution

I tried using x over f, but that would just result in a flat line with a spring constant of 1, which I assume is incorrect. I also solved for k and got k= 2mgh/x^2, but that still results in a straight line of a slope. I'm guessing you have to square height or mass or something but I am very confused. I don't necessarily need a complete answer as much as a guideline. I'd like to figure this out myself.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 13, 2012

### cameronjrhea

I don't know why the 1,2, and 3 showed up again. Sorry. I'm new here

3. Nov 13, 2012

### haruspex

It does? What two variables in that equation would you be plotting as x and y?

4. Nov 13, 2012

### cameronjrhea

x^2 on the x, and 2mgh on the y.
I know that I could put mass or weight on for one variable, and height for the other, but I don't think that would solve for k, right?
Would height/x on one axis and mg on the other solve leave a slope of k? I don't even know how to check if I am right.

Last edited: Nov 13, 2012
5. Nov 13, 2012

### haruspex

But x is constant, right? Only m and h vary, so you need to plot (some function of h) against (some function of m). If you want to extract k as the slope then it needs to be in the form f(h) = k*g(m). See if you can rearrange the equation that way, then plot y=f(h) against g(m).[/QUOTE]

6. Nov 13, 2012

### cameronjrhea

So with the equation 1/2kx^2 = mgh, you could extract out h and get

h=kx^2/2mg

so h/x^2= k/2mg
That way you could plot h/x^2 on one axis and 2mg on the other. and k would be the slope? I think that I understand the physics part pretty well, but I'm struggling with the math.

7. Nov 13, 2012

### haruspex

Not quite. That would be of the form Y(h) = k/X(m). To get a straight line you need to plot Y(h) = k*X(m). So what should the function X(m) be?

8. Nov 13, 2012

### cameronjrhea

1/ 2mg
God this is confusing me way more than it needs to
So that goes on the x axis, and h/x^2 would be on the y axus because h/x^2 would be equal to k * 1/2mg.
Is this correct?
Thanks again for the help!

9. Nov 13, 2012

You got it.