For certain springs we have been taught that Hookes law, $F=-kx$ works. For such springs, what factors determine k. For instance, k is proportional to length. Can k be expressed in terms of dimensions, mass and other fundemental quantities, or is k a fundemental property of the spring itself?
3 Answers
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A slightly more fundamental way of looking at elastic behavior is in terms of the elastic moduli. These still assume a (locally) linear relationship between the extent or shape of an object and the applied force, but they are defined in such a way that they only depend on the material and you have to put it the dependence on length and area.
To use the elastic moduli we must first define stress and strain. I'm going to stick to a simplified definition that will cover your application only.
Stress, $\sigma$, is defined as the applied force divided by the cross-sectional area normal to the force. $$ \sigma = \frac{F}{A} \,.$$
Strain, $e$, is defined as the change in length divided by the unstressed length $$ e = \frac{\Delta L}{L} = \frac{L - L_0}{L_0 \,.}$$
With these definitions in minds the Young's modulus, $E$, which applies to stretching forces, called tension is defined such that $$ E = \frac{\text{stress}}{\text{strain}} = \frac{\sigma}{e} = \frac{F \, L_0}{A \, \Delta L} \,. \tag{*} $$ As with other moduli the Young's modulus is a property of the material the spring is built from and is tabulated.
That's it for the first order behavior of materials under tension.
For some materials, it is a reasonable approximation to use the Young's modulus for compression too, for others you have to look up a different modulus.
You can recover Hooke's Law from this by rearranging (*) to get $$ F = \frac{A E}{L_0} \,\Delta L \,,$$ implying that for stretching springs we have $k = \frac{A\,E}{L_0}$. We immediately see that thicker springs are stiffer and longer springs are less stiff, which hopefully accords with your intuition.
By the way, I cheated a little bit in choosing the Young's Modulus. You see, it's easy to explain because the geometry is simple, however I believe that the response of coil springs (a very common type, afterall) should actually be computed in terms of the Sheer modulus with some extra effort to account for the geometry. Not much fun for a Stack Exchange answer and I'm not sure I could get it right off the top of my head in any case. None the less, the resulting behavior is linear to a first approximation.
answered Jan 28, 2015 at 4:07
dmckee --- ex-moderator kittendmckee --- ex-moderator kitten
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Hooke's law is only a first order approximation of how elastic materials such as springs behave. Hence, $k$ isn't really a fundamental constant but a convenient constant of proportionality for a useful empirical law. The fact that $k$ incidentally is proportional to length doesn't mean that k can be expressed in terms of fundamental quantities, because it isn't really a physical quantity rather than a mathematical constant of proportionality.
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2$\begingroup$ $k$ is inversely proportional to the length! $\endgroup$ Commented Jan 28, 2015 at 3:51
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we know
x = FL/AY
or F= AY/L x
on comparing with F= kx we get k= AY/L thus spring constant dependsupon area , young's modulusand length of the spring
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$\begingroup$ Could you expand this a little bit? FL and AY aren't obvious to people who aren't familiar with the subject matter. $\endgroup$ Commented Feb 28, 2018 at 19:01