# Hooke’s law and the stiffness of springs

- 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.^{© “Hookes-law-springs” 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?

*inversely proportional*to the length of the spring, assuming we are talking about a spring of a particular material and thickness.

*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.

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

*harmonic motion*.

## Answers

\[\Large \frac{10}{m}=\frac{3}{5}.\]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, whereSolving we get \(\normalsize m=50/3=16.7\) kg.\[\Large{\frac{F}{F_1}=\frac{8k}{12k}=\frac{2}{3}}\]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 thatand so \(\normalsize{F_1}\) must be \(\normalsize{\frac{3}{2}}=1.5\) times as large as \(\normalsize{F}\).

#### Maths for Humans: Linear, Quadratic & Inverse Relations

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