Hooke's law and the stiffness of springs
Hooke’s law is a fundamental relation that explains how a weight on a spring stretches that spring. There is a fundamental direct proportionality here, with a constant of proportionality called the spring constant \(\normalsize{k}\). However when we inquire as to the relation between \(\normalsize{k}\) and the length of the spring, we find an inverse proportionality! Hooke’s law explains the oscillation of a spring and the connection with circular functions.
In this step you will learn

about the mathematical form of Hooke’s law

how the spring constant changes with the length of the spring

how the oscillation of a spring is a model for harmonic motion.
Hooke’s law: Restoring force is proportional to displacement
In physics, the motion of a oscillating spring is an important oscillation, that yields the familiar cosine and sine circular functions as amplitudes depending on time. The origins of this motion are in a particularly simple linear relationship that springs have, up to a point, when you displace them.
^{© “Hookeslawsprings” Svjo/Wikimedia Commons CC BY SA}
In the diagram, we see that doubling the force doubles the displacement. This is another way of stating Hooke’s law. This was discovered by the British physicist Robert Hooke in 1660, and stated as
\[\Large{F=kx}\]where \(\normalsize{F}\) is the force required to create a displacement of \(\normalsize{x}\) in the position of a spring. (Sometimes this law appears with a negative sign and the meaning of the force reversed.)
It is only a valid law for relatively small values of \(\normalsize{x}\), but in this range if you want to double the displacement, you need to double the force on the spring. In this case the constant of proportionality \(\normalsize{k}\) depends on the stiffness of the spring.
An example
Suppose we have a spring which is naturally \(\normalsize 6\) cm long when hanging freely. If we place a small weight of \(\normalsize 30\) gm on the spring we notice that it stretches by \(\normalsize 2\) mm. From this we can use a linear relation to predict what kind of extensions we get with other masses. If we double the mass to \(\normalsize 60\) gm we expect the string to stretch twice as far, to \(\normalsize 4\) mm. If on the other hand we replace the mass with \(\normalsize 10\) gm, then we expect the extension to also reduce by to a third of what it was, namely to \(\normalsize 2/3\) mm.
Q1 (E): If a mass of \(\normalsize 10\) kg on a heavy industrial spring of \(\normalsize 100\) cm creates an extension of \(\normalsize 3\) mm, then how much extension would be caused by \(\normalsize 15\) kg? How much mass would we have to load onto the original spring to get an extension of \(\normalsize 5\) mm?
How does the spring constant depend on the length?
But aren’t we discussing inverse relations? Hooke’s law is an example of a direct proportionality, as we discussed in week 1. But behind this law is another law which is a fine example of an inverse proportionality.
Suppose we have a given spring with a given spring constant \(\normalsize{k}\). What happens if we cut this spring into two equally sized pieces? One of these shorter springs will have a new spring constant, which will be \(\normalsize{2k}\). More generally, the spring constant of a spring is inversely proportional to the length of the spring, assuming we are talking about a spring of a particular material and thickness.
So suppose we cut the spring in the example above exactly in two, creating two shorter springs each of length \(\normalsize 3\) cm. One of the smaller springs will have a spring constant which is twice the original. That is because the spring constant and the length of the spring are inversely proportional. That means that the original mass of \(\normalsize 30\) gm will only yield a stretch of \(\normalsize 1\) mm on the shorter spring. The larger the spring constant, the smaller the extension that a given force creates.
Hopefully this makes intuitive sense — it should not be a surprise. If we think of the original spring as being two shorter springs attached together, then the mass of \(\normalsize 30\) is stretching both smaller springs by \(\normalsize 1\) mm, giving a total stretch of \(\normalsize 2\) mm.
For those who are physically inclined, with proper units in this case the force is \(\normalsize{F=0.3}\) Newtons, and the original displacement is \(\normalsize{x=0.002}\) m. So since \(\normalsize{F=kx}\), in these units \(\normalsize k=\frac{0.3}{0.002}=1500\) N/m. And a Newton/metre is really the same as a \(\normalsize \text{kilogram}/\text{sec}^2\).
Q2 (M): If a given force \(\normalsize{F}\) on a spring creates a displacement of \(\normalsize{8}\) cm, and then we cut this spring into three equal pieces, how much force should be applied to one of the pieces to create a displacement of \(\normalsize{4}\) cm?
Spring oscillation and harmonic motion
The particularly simple relation between the restoring force and displacement in Hooke’s law has a lovely consequence for the motion of an oscillating spring. If you pull a weight on a spring down and let it go, then it will oscillate around its mean position in what is called harmonic motion.
Rather surprisingly, this is exactly identical to another fundamental motion: the \(\normalsize{y}\)coordinate of a particle which moves around a unit circle uniformly at a constant speed. If you look at these two examples, hopefully you will see the similarities in motions.
The reason for this seeming coincidence has to do with differential equations, but it comes down to the relatively simple relation of Hooke’s law.
Answers
A1. Since by Hooke’s law the extension is directly proportional to the force, which for a hanging weight is directly proportional to the mass, we argue that if a mass of \(\normalsize 10\) kg creates an extension of \(\normalsize 3\) mm, then a mass of \(\normalsize 15\) kg would cause an extension of \(\normalsize 4.5\) mm. To get an extension of \(\normalsize 5\) mm, we would need a mass of m, where
\[\Large \frac{10}{m}=\frac{3}{5}.\]Solving we get \(\normalsize m=50/3=16.7\) kg.
A2. If the spring constant is originally \(\normalsize{k}\) then by Hooke’s law the force applied is \(\normalsize{F=8k}\). Now for one of the smaller pieces, we know the spring constant has tripled, to \(\normalsize k_1=3k\). If the new force required is \(\normalsize{F_1}\), then \(\normalsize{F_1=4k_1=4(3k)}\). It follows that
\[\Large{\frac{F}{F_1}=\frac{8k}{12k}=\frac{2}{3}}\]and so \(\normalsize{F_1}\) must be \(\normalsize{\frac{3}{2}}=1.5\) times as large as \(\normalsize{F}\).
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