By Kesheav Suraj (30640261)

Welcome everyone to my blog post. In this blog post, I will talk about Hooke’s Law and its application and characteristics. Before I get started, I will give a brief introduction of Hooke’s Law. I hope that you will find this blog post interesting and easy to follow.

What is Hooke’s Law?

“Hooke’s Law is a law credited to Sir Robert Hooke, a contemporary of Sir Isaac Newton.” the law states that the extension of a spring is directly proportional to the applied force. It can be represented as:

F =-kx (where k is the spring constant, and x is the extension of the spring)

As you can see, the above equation represents Hooke’s Law and it can be observed that extension is directly proportional to force. Hence, if the mass suspended was doubled, the force would also be doubled, leading to the extension being doubled. (Hogg, n.d.)

Since each spring is different and unique, each spring has its own spring constant. The spring constant is also known as the stiffness of a spring.

As said earlier, the extension of a spring is directly proportional to the applied force. Therefore, we can tell that the graph of Hooke’s Law appears to be a straight line passing through the origin.

However, “beyond some critical force, the graph ceases to be a straight line. Once the elastic limit is reached, the ratio of extension to applied force becomes highly non linear. There is no guarantee that, when the force is removed, the spring will return to it’s original dimensions.” This means that the behaviour of the spring has become plastic, and with enough force applied, the spring might break. (Hogg, n.d.)

The experiment

The experiment that I had conducted investigates the behaviour of three materials, giving the following set of results. The first two results, y1 and y2, are for two different elastic materials, which are both still in their linear regions. The results, z, describe the behaviour of a material which has gone past its elastic region (in the plastic region). x is the force applied (in Newtons) and y1, y2 and z are the deformation (in mm). The results are to be analysed using MS Excel.


x/N y1/mm y2/mm z/mm
1 3 2.2583 2.375
2 4.5 4.3166 9.375
3 6 6.3749 28.375
4 7.5 8.4332 65.375
5 9 10.4915 126.375
6 10.5 12.5498 217.375
7 13 14.6081 344.375
8 14 16.6664 513.375
9 15 18.7247 730.375

The results for y2 and z were obtained using these formulae (on MS Excel):

y2 = (a+0.5)x + c,

z = x^3 + b,

where a = 1.558, b = 1.875 and c = 0.2

Graph 1 and 2

The graphs of y1 and y2 show that the elastic materials are still in their linear regions. This means that the materials are stretched but they have not extended their elastic limits.

Finding the points where the graph meets:

y1 = 1.5583x + 1.375

y2 = 2.0583x + 0.2

To find where y1 and y2 meet:

y1 = y2

1.5583x + 1.375 = 2.0583x + 0.2

1.175 = 0.5x

x = 2.35

When x = 2.35N,

y = 5.037mm

The lines meet at the point (2.35,5.037)

Graph 3

The graph of z versus x however, differ from the graphs of y1 and y2 in the sense that it is not a straight line but a curve. This is because the elastic material z has been extended over its elastic limit and has become a ‘plastic.’


It is seen in graph y1 that there is an anomaly. This can be due to:

  • parallax error when reading the extension using the ruler
  • not waiting for the spring and mass to reach equilibrium
  • not being able to align the ruler parallel to the spring extension
  • the spring being deformed
  • the measurement devices and masses being inaccurate


Hogg, M., n.d. Mechanical Sciences; Lecture 2 Elasticity, s.l.: s.n.

Khan Academy, n.d. What is Hooke’s Law?. [Online]
Available at:
[Accessed 14 November 2018].

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