# Hooke's Law: Is the extension of a spring the same as its displacement?

This question might make no sense, but I'd like to ask it anyway.

The elastic potential energy of a spring is the area under the force-extension/compression graph.
The work done on a spring is equal to the product of force and displacement.
Since the work done on a spring = elastic potential energy, displacement = extension/compression of spring

Problem is that displacement and extension/compression seem to be different.

^As seen above^, the displacement of the box is from the initial centre of mass to the final centre of mass.

However, ^this image above^ shows the difference between displacement and extension/compression. (x is compression)
Note: The red dot is the centre of mass

1) Is the displacement of a spring the same as its extension?
2) Is the way I am talking about displacement wrong?

By the way, please let me know if there are any difficulties understanding my drawings.

• The position of your red dot (your origin of displacement?) should not be changing. – QuIcKmAtHs Dec 19 '18 at 9:01
• Sorry for the poor labelling of my drawing. The red dot is the centre of mass – helpme Dec 19 '18 at 9:11

You should only consider the displacement of the end of the spring, since this is where the force is applied (the force that compresses/elongates the spring, which is the force that does the work which is stored as potential energy).

If you wish to consider each individual particle of the spring, which will lead you to the red dot movement as an averaged displacement, then you also must count in that the spring constant is gradually different from "particle" to "particle".

Hooke's law combines spring force $$F_{sp}$$ with spring constant $$k$$ and compression/elongation $$\Delta x$$ (for simple idealized springs far from extreme cases):

$$F_{sp}=-k\;\Delta x$$

If you consider the entire compression/elongation, then you can also safely use the "full" spring constant $$k$$.

But $$k$$ depends on not only intrinsic material properties (such as stiffness and elasticity/Young's modulus...) but also on geometry (winding spacing, diameter, thread thickness...). If you consider each individual particle of the spring and want to use Hooke's law on that particle with it's different $$\Delta x$$, then you cannot use the same value for $$k$$. The geometry is different; seen from this particle there is not "the same amount of spring" pushing back/forward on it.

If you do find a useful new value for $$k$$, then you are free to use the different $$\Delta x$$ that fits to that particular particle. I do not think it is useful and worthwhile to try to figure out an "average" $$k$$ to fit the "average" $$\Delta x$$ according to your red dot, but it should be possible.

Well, it ultimately depends. Displacement is the distance from your current point to your point of origin. Extension/Compression refers to the displacement from the equilibrium of the spring.

In your case, let’s say we have a horizontal spring stuck to a wall oscillating horizontally, and you are to take your origin at the wall, then they are different. However, if origin is taken as the equilibrium position of the spring, then the displacement will be the same as the compression/extension.

• Wait, "origin of displacement"? Does this mean that the displacement does not have to be compared from the centre of masses? If so, how can the displacement of the centre of mass/equilibrium point be different even though they should be the same? – helpme Dec 19 '18 at 9:23
• @helpme displacement is usually taken from the point of equilibrium – QuIcKmAtHs Dec 19 '18 at 10:13

Let the spring be orientated parallel to the $$\hat i$$ direction and the displacement of one end of the spring from its natural length position be $$\vec x = x \hat i$$ where $$x$$ is the component (can be positive or negative) of the displacement .

Convention has it that if $$x>0$$ we call that an extension and if $$x<0$$ we call that an compression, however you can call $$x<0$$ a negative extension.
When $$x=0$$ the spring is its natural length.

From this you should gather that the extension $$x$$ (scalar) is the component of the displacement $$\vec x$$ (vector) of one end of the spring in a particular direction which is $$\hat i$$ in my example.

If Hooke's law is obeyed the force that the spring exerts on an external object $$\vec F = - k \vec x \Rightarrow F\hat i = - k \,\hat i \Rightarrow F = - kx$$ where $$k$$ is a property of the spring called the spring constant.

So if $$x$$ is positive the spring is pulling in the negative $$\hat i$$ direction and if $$x$$ is negative the spring is pushing in the positive $$\hat i$$ direction.

The elastic potential energy stored in a spring is $$\frac 12 \,k\,x^2$$ and you will note that this means that the energy stored only depends on the magnitude of the displacement of one end of a spring from its natural length position.
This is not unreasonable as the energy stored in the spring is minus the area under a force against extension graph $$\displaystyle -\int_0^x \vec F \cdot d \vec x$$ is the same if $$x=+2$$ or $$x=-2$$.