What is the Sign Convention for Displacement in Hooke's Law?

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Homework Help Overview

The discussion revolves around the sign convention for displacement in the context of Hooke's Law and simple harmonic motion. Participants are exploring how to define positive and negative displacements when dealing with springs, particularly in scenarios involving compression and extension.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the arbitrary nature of coordinate systems and how this affects the sign of displacement in Hooke's Law. Questions arise regarding whether displacement should be considered positive or negative based on the chosen coordinate system, especially in specific examples involving springs.

Discussion Status

The discussion is active, with participants offering insights into the importance of consistency in coordinate systems. Some have provided examples to illustrate their points, while others express confusion about how to determine the direction of forces based on displacement signs. There is no explicit consensus, but various interpretations are being explored.

Contextual Notes

Participants are navigating the complexities of defining positive and negative displacements in relation to equilibrium positions, and there are references to specific homework problems that illustrate these concepts. The discussion highlights the need for clarity in communication regarding coordinate systems.

Alethia
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I was doing my homeowrk taht dealt with simple harmonic motion and solving problems with Hooke's ewuation (F=-kx). I was a bit confused though on the displacement and when it is negative or positive. For example, when a spring is compressed is the displacement (x) positive or negative? And when a spring is pulled, is the x positive or negative? Yeah, thanks! :smile:
 
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Depends on your coordinate system. You can choose a coordinate system that makes either one true.

The point of the negative sign in Hooke's Law is to make the force act in the opposite direction of the displacement. That is, the force always pushes the block back to the equilibrium (x = 0) position. If x is negative, then the force must act in the positive direction to push it back to x = 0. If x is positive, then the force must actin the negative direction to push it back to x = 0.

cookiemonster
 
It depends on the problem then, I am guessing. Hm, let's see... let's say for example, in this problem:
A fuzzy ball attached to an elastic cord is suspended from a ceiling to be a toy for a cat. As the cat plays, the toy is pulled 15 cm and released. If the toy has a speing constant of 65 N/m, what is the spring force acting on the toy at the moment it is released?
. The displacement would then therefore be negative right, because it is pulled away from equilibrium and must travel up (positive direction) to return to equilibrium. Is that right? :confused: Hehe, thanks again!
 
It's totally, completely, 100% arbitrary. You could just as easily have stated the coordinate system was the opposite what you claimed, i.e. that down was the positive direction and that up was the negative direction. That being said, the coordinate system you did select is by no means incorrect.

The only things that are important are that, #1, you are totally consistent with your coordinate system. Once you select your coordinate system, you must use it throughout the entire problem, start to finish. And #2, you specify which coordinate system you're using so you do not confuse your readers.

cookiemonster
 
Alethia said:
It depends on the problem then, I am guessing. Hm, let's see... let's say for example, in this problem: "A fuzzy ball attached to an elastic cord is suspended from a ceiling to be a toy for a cat. As the cat plays, the toy is pulled 15 cm and released. If the toy has a speing constant of 65 N/m, what is the spring force acting on the toy at the moment it is released? ".
The displacement would then therefore be negative right, because it is pulled away from equilibrium and must travel up (positive direction) to return to equilibrium. Is that right? :confused: Hehe, thanks again!

Yes, you are confused: "negative right"?? Does that mean you are taking negative to mean "to the right" or are just asking "is this right"? Neither one
makes any sense!

There is no "to the right" in this problem, the ball is suspended from the ceiling and then moves up and down.

On the other hand, it makes no sense to ask "Is it right that the displacement would be negative" because the whole point is that the displacement is either positive or negative depending on the direction of motion. Most people would probably choose positive displacement to be up and negative down (so if you meant that the initial displacement [the ball is pulled down 15cm] is negative THEN you are correct) but you could choose either way, as long as you are consistent.
 
LOL when I said "Right" I did mean "is this right?" Not right as in the direction right (oppposite of left). Sorry O was not clear enough.

Sorry but I have another question. XD In this particular problem:
A shopper places some fruit in a spring scale at a supermarket. If the spring has a constant of 420 N/m and is compressed from its equilibrium position by 4.3 cm, what is the spring force on the scale at the moment it is released?
x equals a positive 4.3 cm because I am assumign the scale is pushed down, and when returnign to equilibrium it goes up (along the y axis). Is that correct? XD
 
Alethia said:
Sorry but I have another question. XD In this particular problem:x equals a positive 4.3 cm because I am assumign the scale is pushed down, and when returnign to equilibrium it goes up (along the y axis). Is that correct? XD
As cookiemonster and HallsofIvy explained, what you call a "positive" displacement depends on your arbitrary choice of coordinate system. If you choose down as positive, then the displacement is positive, and the force (-kx) would be negative (otherwise known as "up").

So, your answer is OK. But only because you chose down as positive.

Better not to get hung up on the sign of the displacement. Just find the magnitude of the force and then use your understanding of how a spring works to figure out the direction of the force.
 
But if the direction of the displacement is arbitrary, dependent upon which coordinate system one uses, isn't the answer arbitrary as well? For example, what if someone else chose a different coordinate system than I did in the previous example problem? Wouldn't the final solutions (the force) come out as different values, one being positive and one being negative? Or is that totally irrelevant (the negative signs or what not)? Could you leave out the negative signs in the equationand displacent and go solely on knowledge of force? Because in my textbook it syas that the elastic force is always oppositethe direction of the mass displacement from the equilibirum. I am confused with this concept because if the coordinate system is arbirtray then how does one determine the direction of the force? I don't know if I am making sense... mayeb I'm just thinking to in depth.. but please calrify? Thank you.
 
That's the point of "consistency." While you pick your coordinate system, everything is arbitrary. But once you've picked it, you got to stick with it until the very end, and it's not longer arbitrary. If you decide to change it mid-way through, you got to start all over again.

cookiemonster
 
  • #10
By the way, you say "x equals a positive 4.3 cm because I am assumign the scale is pushed down, and when returnign to equilibrium it goes up (along the y axis). Is that correct?"

Please don't confuse yourself over "x" and "y" axes. This motion is in a straight line. If you are going to call the distance pushed down "x", be sure that you understand (you said it, not me!) that you are measuring that along the y- axis!

It's not really important here whether the "spring" is being compressed horizontally or vertically.

For this problem, all you need to think is "f= -kx" so, since k= 420 N/m and x= 0.043 m (keeping your units straight IS important! since k is in Newtons per meter, x must be in meters also), the force will be -(420)(0.043) Newtons and the "-" just tells you that the force is in the opposite direction to the motion: since the spring is compressed (downward), the force is upward.
 
  • #11
Alethia said:
But if the direction of the displacement is arbitrary, dependent upon which coordinate system one uses, isn't the answer arbitrary as well?
Actually the real direction of the displacement is not arbitrary! Only your description of it as positive or negative is arbitrary. So, yes, if they asked "What is the sign of the displacement?" any answer would be meaningless until you defined what you mean by positive and negative. But, wisely, they didn't ask that did they? :smile:

For example, what if someone else chose a different coordinate system than I did in the previous example problem? Wouldn't the final solutions (the force) come out as different values, one being positive and one being negative?
Sure, but your answer better be that the force is up (I don't care what the arbitrary sign is). (Note: I am assuming that the scale is stretched down from equilibrium, so the force points up.)
Could you leave out the negative signs in the equationand displacent and go solely on knowledge of force?
As long as you can specify the direction of the force, I don't care how you do it.
Because in my textbook it syas that the elastic force is always oppositethe direction of the mass displacement from the equilibirum.
Right! That's what the minus sign in Hooke's law means. If you stretch (or compress) it down the force is up; stretch it left, the force is right. Get it?
 
  • #12
It's amazing how scientists can take a simple idea and make it complicated... :-p
 
  • #13
And now if we say that positive is up, a good bit toward the observer, and just a little to the right...

cookiemonster
 
  • #14
Yeah I get it now, I think I was just over analyxing it instead of taking it as it was. Thanks everyone! Much appreciated.
 

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