# Calculate a spring constant using measurements from a Hooke's Law

1. Oct 31, 2005

### Senor_Blue

I need to calculate a spring constant using measurements from a Hooke's Law Apparatus, a spring, and some weights. Frankly, I have no idea what I need to do. I've taken the measure ments, and discovered the formula F=kx, where F=Force, k=spring constant, and x=the compressed distance.
I'm not sure if this is the proper formula, but heres what I tried to figure out.
If Im solving for k, then k=F/x.
Problem is, I do this for the different lengths of the spring and masses of the weights, and get differentiating answers for the constant.
So here are my questions:
1) Am I even using the right formula? If I'm not, you can go ahead and call me a dunce now and give me the right one.
2) Shouldn't a constant be...well...constant?
And help on this matter would be greatly appreciated.

2. Nov 1, 2005

### Integral

Staff Emeritus
It would help a lot if you were to show us some of your data and your calculations. You are using the rigth formula.

3. Nov 1, 2005

### big man

So is this a spring hung vertically and you are putting on weights??

If so then your force is just the wieght, mg. Did you measure how much each mass displaced the spring from equilibrium? Then all you do is apply your formula that you were using. Remember that the SI units for k are N/m so convert your displacement into metres as well (if they aren't).
There should be a different displacement from equilibrium for each mass you put on, so you should get roughly the same constant. I say roughly because you have to allow for experimental error. But yeah from the information you've given it would seem that the only real mistake you could have made would be in the measurement of the displacement x. You did take the unstretched length of the spring (with no mass on it) as the equilibrium point?? Then did you put each mass on one by one, measuring the displacement from that equilibrium point??

4. Nov 1, 2005

### process

Hello Senor_Blue, yes you are using the correct equation F=-Kx, the negative sign herejust to indicatate the displacement (x) of the spring which is opposite to the spring force (F), The spring force (F) in the equation is the equal to the applied force of course and in your case are the weights you load the spring with. (K) is the Stiffeness of the spring. Its a material property that depends on the spring, a large value of 'K' needs more load to strech/compress a spring than low values of 'K'. As you can see F=-Kx is the equation of a straight line, so (K) is the slope. Plot your values that you have on a graph and then calculate the slop along any part of the line (Y2-Y1/X2-X1) and this value is you 'k'. Notice it has to be a line since you are dealing with the elastic part of the behaviour of the spring!

5. Nov 1, 2005

### dfx

Hi, I am doing exactly the same thing and I have 2 questions:

1. Would I be right in saying that because I am interested in the scalar quantity k (spring constant - I'm finding out from the investigation), the negative sign on the -x can be ignored?

2. I've plotted a graph of F against x and the evaluated the gradient k to be the spring constant. However, my procedure involves changing masses and measuring the corresponding extension - which means that the force is the independent variable and extension is the dependent variable, I think. Thus, strictly speaking, would I have to plot a graph of x against F and determine 1/k to be the spring constant, or would it not matter? Testing several times shows exactly the same answer, but does it matter?

6. Nov 1, 2005

### ZapperZ

Staff Emeritus
Practically, yes. All the negative sign is saying is that the direction of the restoring force (F), and the direction of the INCREASE in displacement (x) are in the opposite direction to each other. In extracting the value of k, this direction issue is irrelevant.

It does not matter. You could have easily asked for the amount of mass needed to get an extension of so-and-so, and the spring-mass system won't know any better.

Zz.

7. Nov 2, 2005

### dfx

awesome.. thanks ever so much.

8. Nov 2, 2005

### Spv

Is there any experiment to accurately work out the constant of a spring, other than hanging it vertically, adding weights to it and then using k=F/x, if you have weights, a spring, an optical pin, an anti-parallax mirror, a peice of wire and a ruler?

Last edited: Nov 3, 2005
9. Dec 28, 2005

### dfx

Use Simple Harmonic Motion.