# I Hooke’s law -- Only valid for small delta-x values?

1. Nov 1, 2018

### Arman777

Last edited by a moderator: Nov 2, 2018
2. Nov 1, 2018

### ZapperZ

Staff Emeritus
This is incorrect.

As long as the spring isn't stretched beyond its elastic limit, then Hooke's law is still valid. Have you ever done Hooke's law experiment before? It doesn't say that the stretch has to be small.

You may be confusing this with pendulum oscillation.

Zz.

3. Nov 1, 2018

### Arman777

I remember that our teacher said like that. But now I am thinking that, he might said that for the oscillatory motion for a mass attached to the spring. For such motion we need small displacement. I definitly remember it because I wrote it on my notebook :)

If this is the case and OP asked for the frequency So that why I feel like to say that it should have small displacement for observe such frequency.

Other cases when we attached a mass to a spring as you said hook's law is still valid until spring's elastic limit is exceeded.

4. Nov 1, 2018

### ZapperZ

Staff Emeritus
But there's nothing in the physics that requires this, as long as F is linear with x.

Again, have you done the experiment? If you have, how far did you stretch the spring? Can the graph of F versus x be represented by a straight line? If it can, then that entire range of x can be used during oscillation.

Now, there may have been other non-physics reason why your teacher do not want you to have the mass oscillate with large amplitudes, especially if the mass is simply resting on a hook. A large oscillation may cause the mass to dislodge off the hook. But this has nothing to do with the physics.

Zz.

5. Nov 2, 2018

### Arman777

So Hooke's Law for valid for small x ? Can someone please explain it to me. . When we attached a mass to spring. At some point the spring reaches its equilibrium position. After that point are you saying that Hooke's Law is valid for small x. Can you also explain why ?

I thought it yesterday, and my idea is that when we apply large force to give large displacement we are creating an external force. And if this force gets larger then the equation of the motion will change due to this external force. So we need to keep as small as possible.

Last edited: Nov 2, 2018
6. Nov 2, 2018

### Staff: Mentor

This relates to the material behavior rather than to Newton's 2nd law. Not all materials exhibit a linear stress-strain curve, even in the region of elasticity. But, for most metals, the behavior will be linear up to the point where the metal begins to yield. For non-metals, such as rubber or plastic, that will not be the case, and a spring made out of a hard rubber (or plastic) will exhibit a non-linear Hooke's law relationship.

7. Nov 2, 2018

### Arman777

I didnt not do the experiment, no. Also thanks for creating this thread :) .

8. Nov 2, 2018

### ZapperZ

Staff Emeritus
Somehow, what "your teacher" has said seemed to be stuck in your head without you understanding anything.

Here's a sample of a Hooke's law experiment. Look at the sample data. The extension length goes to beyond 0.6 m! Is this "small x" to you? Yet, look at the linear relationship between F and x. Are you saying that Hooke's law does not apply to such length?

Zz.

9. Nov 2, 2018

### Arman777

Lol yes :) He said this is important too. I dont know, I thought that is the truth.
Okay I understand it thanks.

10. Nov 2, 2018

### Stephen Tashi

In the experiments with springs that I remember, Hooke's law doesn't work for very small changes in some springs. For the type of spring where the coils of the spring touch each other where there is no tension, it's possible to apply a small tension and not get a measurable change in the spring's length. Once you apply enough tension so the coils begin to separate, Hooke's law begins to work.

11. Nov 5, 2018

### sophiecentaur

Here's another point.
The reason that a coil spring follows Hooke's Law over a wide range is because of its construction. It basically works on the torque on the wire, the thickness (cross sectional area) changes hardly at all until the helix is almost straight. If you just take a straight wire and stretch it, every fractional increase in length is accompanied by a corresponding decrease in the cross sectional area. So the modulus of the material remains the same but the stress increases more as the strain increases . This is nothing to do with the Elastic Limit because the wire can return to its original dimensions but the Strain / Load graph is not a straight line (over anything but small amounts of strain.

12. Nov 5, 2018

### Staff: Mentor

Yes. This is a good point. A spring stretches by relative shear of cross sections (combined with the helix geometry), not by extension along the spring contour.

13. Nov 8, 2018

### Arman777

I guees I find why my teacher said like that (Its kind of late I know)

14. Nov 8, 2018

### PeroK

That's a different matter. That's effectively saying that any restoring force exhibits SHM about an equilibrium point, for small enough displacements. That does require small displacements, in general, so that the higher order terms can be neglected.

If the force is linear in the first place, of course, then it applies for larger displacements.

15. Nov 8, 2018

### ZapperZ

Staff Emeritus
It is why we dislike it when a "reference" is simply a word-of-mouth. It is often unreliable and inaccurate! Just think of the amount of wasted time and effort many of us had to do in just trying to correct you. And this all started in a separate thread with a separate issue!

Zz.

16. Nov 8, 2018

### Mister T

If you have a graph of $F$ versus $x$, such that $F=kx$ then the graph is a straight line with a slope of $k$. Hooke's Law is the assertion that the graph is a straight line. When you measure values of $F$ and $x$ and plot them, you may instead get a curve, in which case we'd say that Hooke's Law is not valid. However, if you zoom in on a small enough piece of that graph, you may approximate it as a straight line, and then you can say that Hooke's Law is valid over that small range of values of $x$.

17. Nov 8, 2018

### Arman777

Its an approximation yes. Its valid for small x but for large x its approximately true.

18. Nov 8, 2018

### Mister T

It's always an approximation, it's just a question of how much.

19. Nov 14, 2018

### DanielMB

Hooke Law

A system when it is deformed in a direction x, reacts to the deformation opposing a restoration force F(x) , this force, most of the time in Nature is a smooth function that could be diferentiated and expressed in Taylor Series around the equilibrium point (i.e. x = 0) as :

F(x) = F(0) + x. F’(0) + (1/2).x^2. F’’(0) + … [1]

In the equilibrium point x = 0 implies that F(0) = 0, therefore

F(x) = x. F’(0) + (1/2).x^2. F’’(0) + … [2]

In the region where the linear approximation is valid, the following expresion will be a valid law

F(x) = x. F’(0) [3]

Making

k = F’(0)

The constant k is a property of the reacting system (i.e. spring constant), is a physical constant

Then

F(x) = k . x [4] this is the mathematical expression of the Hooke Law (HL)

A system will react to a deformation (x) imposed by an external agent, with a restoration force (F) directly proportional to the deformation (x)

Where the Hooke Law is valid, is valid also the Superposition Principle

Superposition Principle, Validity

The Superposition Principle (SP) is so obvious that it is important to clarify that it is not always valid

The SP fails when perturbations in the system exceeds the deformations allowed by the Hooke Law

In the field of sound waves, violent explosions (like a thunder) create shock waves that behave differently than low intensity sound waves, the law that shock waves obey is approximately cuadratic (not linear) and do not fulfil SP

To obtain the limits of validity of the HL for a system, from equation [2] should be …

x. F’(0) >> (1/2). x^2

Simplifying we obtain what it is called the Characteristic Hooke Length (L) for the system

L = 2F’(0)/F’’(0) [5]

Finally

L = 2k/F’’(0) [6]

The HL and SP will be valid for the system when the deformation (x) be

X << L [7]

This is the correct form to express the vague term “a little deformation around the equilibrium point”. Note that [7] is a relative term, relating x & L

When [7] is valid, the system behaves as a linear system

Considerations

There are two reasons whereby a restoration force, derived from a Potential Energy (U), fails in offering a linear response in the proximity of the equilibrium point:

1) The Potential Energy does not vary smoothly enough around the equilibrium point, in a way that the first and second derivatives are not well defined around the equilibrium point, this reason makes the Taylor Series not valid

2) Still when first and second derivatives exist, it could happen that U’’(0) = 0, in this case, to get a stable system, should be accomplished that U’’’(0) = 0, otherwise the system will be unstable, so a small stimulus will produce a big deformation that could damage the system

These reasons are rare in Nature

20. Nov 14, 2018

### DanielMB

I should have written

x. F’(0) >> (1/2). x^2 . F''(0) instead of : x. F’(0) >> (1/2). x^2