Hooke's Law

[url=https://pixabay.com/en/spring-helical-metal-steel-1453075/]"Coil Spring"[/url] by Skitterphoto is in the [url=http://creativecommons.org/publicdomain/zero/1.0/]Public Domain, CC0[/url][br]A coil spring that follows Hooke's law.[br]
Robert Hooke (1635-1703) discovered the law of elasticity of solids. The law is generally called Hooke's law these days in commemoration of his work.  Hooke's law states that springs (or solids in general), when stretched or compressed, will exert a [b]linear restoring force[/b] that tries to restore equilibrium. This simply means that when you stretch (or compress) a spring it pulls (or pushes) back.  Further, Hooke's law states that this restoring force is linearly dependent on displacement from equilibrium.  Such a linear restoring force can be written mathematically as [math]F_x=-k(x-x_0)[/math], where the constant [i]k[/i] is referred to as the [b]spring constant[/b] or[b] elastic constant[/b].  [color=#1e84cc][br][br][color=#000000]Please notice that "linear" in this context has nothing to do with the direction of the motion. It is a statement about the force depending on displacement from equilibrium such that it may be written as a linear function, as seen above. [/color] [/color][br][br]By inspecting the terms in the equation above, it is clear that the elastic constant has units of N/m since force is in newtons and the displacement in meters.  The larger the value of [i]k[/i], the stiffer the spring.  While most springs never get stretched by a full meter from equilibrium, the constant k tells us how much force would be required to stretch a given spring by a full meter (if it could do so without breaking). Since Hooke's law is linear, we know immediately that if k=100 N/m, then to stretch the spring by just 1 cm (1/100th of a meter) it would take only 1N. [br][br]The term [math]x_0[/math] represents equilibrium length of the spring, and x represents the current length, whatever it may be due to stretching or compressing.  There is a minus sign on the right side of the equation since the force opposes the displacement.  If you stretch a spring to the right, the restoring forces pulls back to the left since [math]x>x_0.[/math]  If you compress the spring [math]x<x_0[/math] so that the force is in the positive direction. Without the minus sign the equation would imply that after you pull the end to the right the spring would help you pull even more to the right and vice versa. That would be interesting, but it's not the way nature operates.
Restoring Forces and Stetching Springs
Notice that when we stretch a spring with an applied force, that we are not producing the restoring force. Rather we are opposing the restoring force with our applied force, and are therefore either stretching or compressing the spring away from its equilibrium position. [br][br]If we assume the spring starts at equilibrium x=x[sub]o[/sub] (in which case it is not exerting a restoring force) and we apply an external force F while attempting to stretch the spring, the spring will begin to lengthen. As it does, it will begin to oppose my applied force with a restoring force which will act in efforts to restore equilibrium length. The longer the spring gets, the harder it will fight. There will come a time when the length is such that the restoring force matches my applied force and motion will cease. [br][br]So the rule is simple. If we apply a force and perhaps wait a moment for any vibration or oscillation to die off, the amount of stretch we get is the same one we'd get if we plug in our applied force into Hooke's law in place of the restoring force (since it's the same magnitude). The only difference is that the minus sign shouldn't be there since the forces are equal and opposite. [br][br]If you recall Newton's 2nd law, you might be thinking about the spring's acceleration, and once it's in the act of stretching you might also be thinking of its momentum. All sorts of complications can arise if you do this, and at least at this level we avoid those complications by assuming we are dealing with mass-less springs. The idea is that whatever is attached to the end of the spring contains the vast majority of the mass, so we ignore the mass in the actual coils of the spring. In most cases it's a very good approximation.
Applications of Hooke's Law
While Hooke's law is often seen as a law for springs, we saw in the last section that all solids are elastic and springy like springs.  In fact, Hooke's law generally applies to stretching of objects, compressing of objects, bending of objects and twisting of objects - as long as they are not stretched, compressed, bent or twisted too far.  [br][br]For instance, consider standing on the end of a diving board.  If another person your same weight joined you on the end of the diving board and therefore together you exerted twice the force on the end of the board, the board would bend twice as far.  [br][br]As another example, take something like a meter stick in your hands.  With one hand on each end, twist them in opposite directions.  Hooke's law applies to the force of your twist (or torque) as compared with the angle that you manage to twist the stick.  It will be a linear restoring force as well.  Such twists can be applied in torsion pendula. [br][br] As a last example, it turns out that people who model molecular motions in what are called [b]molecular dynamics simulations[/b], will model the forces between individual atoms with Hooke's law.  As complicated as the quantum mechanics is that governs inter-atomic forces, they can be modeled quite well as if the atoms are little balls of mass connected by bonds which are little springs!  Of course in some sense we should expect this to be true of atoms.  Why?  Because springs and other solids are made of atoms.  Where else would they have gotten their elasticity if not from their atomic makeup?
[color=#1e84cc][b]EXAMPLE[br]The elastic constant of a diatomic nitrogen molecule is [/b][math]k=22.4\frac{mdyn}{\text\AA}[/math][b].  The units commonly used for molecular-sized systems are milli-dynes per Angstrom, where  [math]1dyne =1\times 10^{-5}N[/math] and [/b][math]1\text\AA=1\times10^{-10}m[/math][b].  What is this elastic constant in N/m, which is more commonly used for macroscopic springs?[br][br]If you do the conversion, it should lead to 2240 N/m.  This is actually quite stiff!  And this is a single diatomic molecule of nitrogen. It is an unexpected fact of nature that single molecules are as stiff as macroscopic objects made out of those molecules![/b] [/color]  

Information: Hooke's Law