Some question about Spring force and Deformation

[Jenny_wui]

I have some basic question about spring force. The anchor position for spring is its equlibrium position. As the spring is stretched, when released, it is expected that it should come back to its equilibrim position. But I found when less force is applied, the new equilibrium position is a bit far from the original position of rest. The more force is applied, it comes closer to the original equilibrium position. Could anyone explain this behaviour with mathematics. For a certain amount force if I always want the spring to be back in its original equilibrium position, what do I need to do?
Thanks every body in advance.

 

[Thaakisfox]

well could you give a bit more detailed description of what you did exactly? ws the spring hanging, and moving vertically? or was it on a horizontal table?
if it was on a horizontal table then most probably since you applied less force the friction on the table slowed it down.
But share your experiment with a bit more detail, and then I could help more.

 

[alxm]

Well, the simplest model of the force of a spring is Hooke's law, that the force is proportional to the displacement (stretching) of the spring. Hooke's law is usually only true for small displacements. -Springs cannot be stretched to any length or compressed to any length, the begin to deform, etc. A Hooke's law spring will continue to oscillate forever (harmonic oscillator), since there's no friction. The simplest way to include friction is to create a 'damped oscillator', where you have a term that represents friction, air-resistance, or whatever losses you may have, as giving rise to a force proportional to velocity. A damped oscillator will also return to its original position.

So to describe what you're seeing, you'd need a more accurate description of how the force on the string changes with displacement, and a more accurate description of how the friction/damping acts on the spring. You could model this in many ways, perhaps doing an expansion of spring's force in higher powers of x (with corresponding spring constants), and the friction in higher powers of v (and friction terms).

If you then balance the forces, as one does in deriving the equations for the harmonic/damped oscillators, you get an ordinary differential equation. But unlike the simpler models, there likely isn't an analytical solution to it. It'd be a non-linear ODE. One of the characteristics of these is that they often have many different equilibrium points, in other words, depending on how the system starts out (and it may be extremely sensitive to this, which is part of chaos theory), it can converge to different stationary points. So it's not entirely surprising that it may return to different points if compressed and released at different distances.

That's basically the best "mathematical" explanation I can give you, short of doing actual calculations. Basically, you've called our bluff. The reason why damped oscillators are taught in all the textbooks isn't because it's a great model of real-world springs, but because it's a model that happens to have a straightforward exact solution. But you see, as soon as you step off the path of neat idealizations and exactly-solvable models, even the simplest things can get quite complicated. Welcome to physics (and the forums)!

 

[Jenny_wui]

Thank you very much. Actually I am doing some programming in graphics. Here I have chosen some anchor point. As the distance from the anchor point increases, force increases and displacement takes place proportionately. But when in released position, it always stops a bit away from the anchor position. I am not doing any experiment. Just want to show something with programming. But I need to show that after release, it comes back to the original anchor position. How to fill in the gap? Could anyone suggest?

 

[PJC01]

The behavior you are describing is known as "spring hysteresis," where the equilibrium position of a spring changes depending on the amount of force applied. This is due to the material properties of the spring, specifically its stiffness or spring constant. The spring constant determines how much force is needed to stretch or compress the spring a certain distance.

When a spring is stretched or compressed, it experiences a restoring force that is proportional to the displacement from its equilibrium position. This is described by Hooke's Law, F = -kx, where F is the force, k is the spring constant, and x is the displacement from equilibrium.

In your case, when less force is applied, the displacement is smaller and therefore the restoring force is smaller. This means that the spring will not return to its original equilibrium position, but instead settle at a new equilibrium position that is further away from the original position. As more force is applied, the displacement and restoring force increase, causing the spring to return closer to its original position.

To always have the spring return to its original equilibrium position, you would need to apply a force that is equal to or greater than the force that caused the displacement. This can be calculated using Hooke's Law and the known spring constant. Alternatively, you could adjust the spring constant by changing the material or dimensions of the spring to better suit your desired behavior.

I hope this helps to explain the behavior of springs and their equilibrium positions. Keep in mind that this is a simplified explanation and there may be other factors at play, such as friction or non-linear behavior of the spring. If you want a more detailed understanding, I suggest looking into the field of elasticity and studying the stress-strain relationship of materials.