# Spring constants and Balances

1. Nov 30, 2007

### Dodirama

Hi all,

I'm a physics teacher at an australian high school, and some kids that I teach have give me some interesting data reguarding spring constants.

They made a contraption to project marbles as part of an analysis of projectile motion, and part of the contraption was a spring, used to apply a force to the marble.

Too late, they realised they would like to know the spring constant of the spring, and were unable to hang masses etc on it ( was in contraption) so decided to measure the force needed to make the spring compress a certain distance using a spring balance to pull the spring into a compressed state.

The data they collected showed a nigh perfect F proportional to x SQUARED relationship.

Could the 2nd spring (in the balance) have affected these results?

Diana

2. Dec 1, 2007

### rdt2

A second spring in series won't make a difference, even if it has a different stiffness. I assume you mean 'stretch' rather than 'compress'? If so, it's most likely that the first spring has been stretched beyond the point where its response is linear. I'm surprised that it didn't hasn't been permanently deformed - or has it?

3. Dec 1, 2007

### Dodirama

No, actually, compress. I was under the impression that hooke's kaw applied in this case too. Perhaps it doesn't?

so, they were pulling... basically a piston? With a spring balance, that caused the spring to compress.

4. Dec 1, 2007

### ZapperZ

Staff Emeritus
In the simplest case, Hooke's law should also apply to compression.

I don't quite understand the "spring balance". How exactly is this done to measure the force of the original spring?

Zz.

5. Dec 1, 2007

### Dodirama

I wish I could draw a picture. Do you know what a spring balance is?

it's a plastic cylinder with a spring in it with a hook on the end. you put stuff on the hook and pull. A scale written up the side gives you readings in either grams (this one was in grams) or newtons.

Their cannon was a hollow metal pipe with a metal rod through the middle that could be pulled to compress a spring, and then be released to let go of that energy.

They hooked the spring balance to the end of the rod, and pulled the rod a certain distance with the spring balance. They recorded this distance and the reading on the balance.

6. Dec 1, 2007

### ZapperZ

Staff Emeritus
OK... so one can get the spring constant of the spring balance, no?

So I don't understand the relationship between the spring balance and the original spring used in the experiment.

Zz.

7. Dec 1, 2007

### Dodirama

we're using the balance to apply a force

8. Dec 1, 2007

### ZapperZ

Staff Emeritus
Let me try to guess here. You attach the spring balance to the original spring, and then you pull on the spring balance? In other words, you essentially have 2 springs in series, and the combination of these two is the displacement that you measure whenever you apply a force?

Zz.

9. Dec 1, 2007

### Q_Goest

Hi Diana,
I'm assuming this is a coiled spring, either extension or compression. A simple coiled spring will of course have a linear spring coefficient as you no doubt already know. The equation to determine spring constant is given by equation 2 here:

This site gives you everything you need to calculate the spring constant. If you have any questions about how to calculate, feel free post those questions here.

The question is, if this is a simple coiled spring, why would you measure a spring constant that isn't linear? Couple of things come to mind:
1. The distance the measuring device is moved through is not linearly proportional to the distance the spring is compressed or extended. This could be because of some lever arm or other mechanism between your measuring devivce and spring.
2. The spring is buckling and therefore is either rubbing on something and friction is coming into play, or as the spring buckles, the distance the measuring device is being moved through is not linearly proportional to the distance the spring is compressed. Note that if the spring is leaning over or bending and not compressing in a nice axial way, your measured k value won't be linear.
3. I seriously doubt the spring material is yielding, but that's possible. Most compression springs are designed such that they don't yield even when compressed to their solid height. And if the spring comes back to its original shape when the force is released, then it obviously hasn't deformed and this possibility is out.

Hope that helps,
Dave.

10. Dec 1, 2007

### Staff: Mentor

The other possibility is that the launcher's spring is "non-Hookeian" in compression. I mean, Hooke's law is not really a law in the sense that it is a fundamental feature of the universe. It is more like Ohm's law where some behaviour is observed to be more or less linear for certain materials or devices. You shouldn't be too worried about violations of Hooke's "law".

11. Dec 1, 2007

### Dodirama

hmmm... good point.

So I wonder what that would mean about the EPE of the spring?

12. Dec 1, 2007

### Staff: Mentor

Just integrate the measured force over the measured compression distance to get the PE as a function of the distance. You will have some numerical and measurement errors, but overall it may be more accurate than relying on Hooke's law if your spring deviates significantly. Since your force seems proportional to x^2 your PE should be proportional to x^3. Plus, it sounds like you already have the data you need.

13. Dec 1, 2007

### TVP45

Was the spring cylindrical? Or perhaps tapered?

14. Dec 2, 2007

### Q_Goest

Hi DaleSpam,
As you say, Hooke’s Law is dependant on the interactions of the material’s atomic interactions. So yes, it’s a property which is fully reducible to quantum level interactions. However, the law holds for all metals (that common springs are made out of) up to the yield point of the material. Stress and strain will always be linearly proportional for a metal spring.

There are however springs made which are not linear as TVP45 eludes to. Springs which are progressively wound for example, have coils which flatten out against each other as the spring is compressed so that the material in the coil can’t be compressed any further. Such springs would look like those on this page:
http://www.wilbersusa.com/images/Spring_progressive250.jpg [Broken]

Last edited by a moderator: May 3, 2017
15. Dec 2, 2007

### TVP45

Yes, that's what I was thinking. A lot of springs in everyday have a
k(x) spring constant, or perhaps just several spring constants. Many motorcycle suspensions, valve springs, etc. are like that. I think they're hard to buy from a standard parts house, but who knows where the students might have picked one up.

Last edited by a moderator: May 3, 2017
16. Dec 2, 2007

### Dodirama

Great- thanks everyone. The kids in question haven't actually even done much differential calculus, so I guess they can be forgiven for not knowing that EPE is the area under a Fx graph...

THOUGH I think we have talked about work being the area under a Fs graph... so, ok, yeah, they could have worked that out - but they wouldn't have had the maths tools to actually work out the integral.

Anyway, they have done much worse things in their write ups... tried to fit a parapola to a sin function that they KNEW was a sin function to find the range of a projectile. Man I'm getting frustrated with it.

Cheers for all your help in satisfying my curiosity.

Diana

Last edited: Dec 2, 2007
17. Dec 3, 2007

### Staff: Mentor

I am sorry, but the OP's data disagrees. This particular spring is decidedly non-linear. Unless you believe that the OP made a measurement error there is no point in asserting the theory when it clearly fails to explain the data.

As you described, you can make springs that are non-linear, even if the metal never yields. In general, Hooke's law does not necessarily apply to a structure as a whole even if it does apply to the metal at every point in the structure.

Last edited: Dec 3, 2007
18. Dec 3, 2007

### Staff: Mentor

They probably could only use the trapezoid rule to approximate the integral. But that probably wouldn't have occured to them anyway if they don't have any calculus background. It is probably not reasonable to expect students in a non-calculus physics course to recognize a potential failure of a "law", understand that the equations they have don't apply, figure out the more general ideas that do apply, and approximate them with unknown math.

19. Dec 3, 2007

### Q_Goest

Hi DaleSpam,
It sounds as if you’re confusing Hooke’s law with the concept that a spring might produce a force which is linearly proportional to the distance moved (i.e.: has a single spring constant, k).

Hooke’s law has nothing to do with the structure as a whole, the elastic modulus is an intensive material property just like density. Hooke’s law only regards the stress/strain relationship of a material, it has nothing to do with springs except that modulus is used to calculate a spring constant. From E. P. Popov, “Mechanics of Materials” second edition:

The applicable pages are also attached for further exam.

Springs can be made with a non-linear force to distance relationship, but the metal they are made from always obeys Hooke’s law. If you still don’t believe this, please reference a reputable source.

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Last edited: Dec 3, 2007
20. Dec 3, 2007

### Dodirama

sounds like you guys are actually on the same page.