Confusion Regarding Hookes law

I am used to seeing Hookes law in the form: $$F = kx$$ but in another one of my books it gives the equation: $$\\ F = \frac{\lambda x}{l}$$ where $$l$$ is the unstretched length, $$x$$ is the extension and $$\lambda$$ is the modulus of elasticity.

My confusion lies in what the modulus of elasticity actually is as I initially thought it was young's modulus but then realised its units are newtons. I don't understand what it is actually measuring as its not something like 'newtons per [unit]'. I would understand if the modulus was $$\frac{N}{l}$$ as that seems to make sense but just the $$\lambda$$ on its own does not Any help?

Physicists tend to use spring constant $$k$$ and mathematicians modulus of elasticity $$\lambda$$.

Suppose that there is a specification for a type of spring comprising the material of which it is made, the diameter of the wire of which the spring is made, the diameter of the helix etc then all springs with this specification and irrespective of their length will have the same modulus of elasticity $$\lambda$$ but the spring constant $$k$$ will depend on the length of the spring.

Suppose that a spring of length $$L$$ extends by $$y$$ when a force $$F$$ is applied. The spring constant of this spring is $$k_1= \frac {F}{y}$$.
The same is true of another spring of the same length with the same specification.

Connecting the two springs in series with total length $$2L$$ means that when a force $$F$$ is applied to the springs each spring extends by the same amount $$y$$ and so the total extension of the two spring of length is $$2y$$.
So the spring constant of this longer spring is $$k_2=\frac{F}{2y}$$ which is half the spring constant of one spring ie $$k_2=\frac 12 k_1$$.

Now consider the modulus of elasticity.
For one spring $$\lambda_1= \frac{FL}{y}$$ and for two springs in series $$\lambda_2= \frac{F2L}{2y}= \frac{FL}{y} =\lambda_1$$.
Thus the modulus of elasticity does not depend on the length of the spring.

Young’s modulus is somewhat different as it is a property the material from which the spring is made irrespective of the size and shape of the material.

If you glue two elastic bars together end to end and apply a lengthwise force to them, the force will make each of them contract or compress according to Hooke's law. The total change in length of the aggregate bar is the sum of the change of length of each of them, so the compound bar follows Hooke's law with a $$1/k$$ that is the sum of the $$1/k$$s of the individual bars.

Conversely, if you take a uniform bar and cut it into two identical pieces, the $$k$$ of each piece must (by symmetry) be twice the original $$k$$.

By extension of these arguments, for a particular combination of material, cross section, etc. the $$k$$ of a bar is inversely proportional to its length. The proportionality constant is the modulus of elasticity you find in the second equation.

It is the same as $$k\cdot l$$, so its unit is $$\rm (N/m)\cdot m$$, where the meters cancel out and leave only newton. Intuitively, units of force correspond to thinking of it as "if Hooke's law worked for arbitrary large displacements (which it doesn't), how much force would be necessary to compress a bar of this material and thickness to length $$0$$?"

[This doesn't work if the bar is so long that it starts to buckle under compression, or so short that non-uniformity near its ends begin to matter -- for example, friction with whatever we use to push on the ends might prevent the material from expanding in the lateral direction when we compress it, so it may appear stiffer than it ought to be according to its modulus of elasticity. But it's a valid approximation for a useful range of lengths].

• Thanks! This answer along with another thing I read about deriving it from youngs modulus helped a lot. – DevinJC Jan 31 '19 at 21:51
• I think you've made a slip in the first para. $k$ as usually defined is halved if you put 2 wires or springs in series. – Philip Wood Jan 31 '19 at 21:53
• @PhilipWood: Indeed, I had $k$ confused with $k^{-1}$ in much of my thinking. I hope it is more correct now. – hmakholm left over Monica Jan 31 '19 at 22:02

$$\lambda$$ is proportional to Young's modulus times the cross sectional area of the spring wire. So, it has units of force.

Actually, when a spring extends, it is because of relative shear rotations of the wire cross sections which couple with the spring helix shape to translate into an axial extension. So it is actually the shear modulus of the metal (which is proportional to Young's modulus) that determines spring axial stiffness.