# Introduction to Hooke’s Law

Hooke’s Law states that the extension of an elastic object is directly proportional to the force applied to it up to the limit of proportionality (elastic limit).

Hooke’s Law can be represented by this equation:

**F = k x**

**F = k x**

is the force applied in Newtons (N)**F**is the spring constant in newtons per metre (N/m), the value for**k**can only be determined experimentally**k**is the extension of the spring in metres (m)**x**

The equation is only valid until the limit of proportionality. If the spring is stretched beyond the limit, it will start to have a plastic behaviour and won’t return to its original length when the weight is removed.

# The Experiment

In the experiment that I have conducted, I investigated the behaviour of three different materials in relation to the force applied on them.

*The apparatus used for the experiment were set up as such:*

I started with a small mass and increased the mass used, measuring and recording the deformation (extension) of the material as I go. I used the same value of mass for all three materials to ensure consistency in my results.The results of the experiment were recorded in the table below.

# Data Collected

Table 1: Values obtained from the experiment

* y1 *and

*represent the deformation for two different elastic materials, which are both still in their linear regions; whereas*

**y2****represents the deformation for the third elastic material, which has gone past its limit of proportionality.**

*z*From the values in *table 1*, graphs can be plotted to show the relationship between the deformations (** y1, y2, z**) against the force applied (

*).*

**x**

# Graphs & Analysis

Figure 1: Graph of

andy1againsty2x

For the graph in *figure 1*, a linear line of best fit can be drawn and an equation in the form of ** y=ax+b** can be derived. I have done so using Microsoft Excel and I had determined the equations for

*and*

**y1****to be**

*y2**and*

**y1=1.5583x+1.375****respectively.**

*y2=2.0583x+0.2*Based on the graph the equations of the lines, it is evident that the deformation of the materials is proportional to the force applied as one increases with the other. However, the variables are not directly proportional as the graphs do not intersect the origin, (0,0). This can be seen as the value of “b” for the equation of * y1 *and

**are**

*y2**and*

**1.375***respectively instead of zero. This may be the result of systematic errors such as:*

**0.2**- Using a material that is already deformed
- Using faulty measuring equipment
- Incorrect method of measuring (human error)

Furthermore, there is also an anomalous point for the graph of ** y1 **at x=7. This could be the result of random error or human error.

Upon further analysis, it can be seen that the gradient for ** y2 **is greater than that of y1 as the value of “a”for the equation of

*is larger. Based on this observation, it can be concluded that the second material has a smaller spring constant and is stiffer than material 1. Thus, making it easier to stretch.*

**y2**In addition to that, it can also be seen that both graphs intersect at one point, meaning that at that point, both materials have the same deformation for the same force applied.

By using the graph from *figure 1*, I have estimated the approximate x-value of the intersection point to be x=2.3N. When I tried to confirm this estimation through calculations, I managed to obtain the value of x=2.35N.

### Working for the calculation:

I used the matrix method on excel to solve the simultaneous equations.

eqn 1 : **y = 1.5586 x + 1.375 **

eqn 2: *y = 2.0583 x + 0.2*

First, I rewrote the equations in the form of a matrix as shown above, then I solved them using the “MINVERSE” and “MMULT” functions in Microsoft Excel, which is equivalent to solvinig by doing this:

AX = B

(A^-1)AX = (A^-1)B

X = (A^-1)B

X = 2.35N

Figure 2: Graph of z against x

For the graph in *figure 2*, a curve can be drawn and a third-degree polynomial equation can be derived. Once again, this was done using Microsoft Excel. The equation obtained for the graph is* z=x^3-2(10^-12)x^2+4(10^-12)x+1.375* , which can be simplified to

*as the values of*

**z=x^3+1.375***and*

**-2(10^-12)x^2****can be considered negligibly small.**

*4(10^-12)x*Based on this graph, it can be seen that the material does not obey Hooke’s Law as the line is not linear and the variables are not directly proportional to one another. This is because the material has already been stretched pass its elastic limit. Therefore, the third material will no longer return to its original shape after the weight is removed and it will remain permanently deformed (plastic behaviour).

# Conclusion

Ignoring the systematic errors done, the first and second materials can be said to obey Hooke’s Law as the force applied is directly proportional to the deformation of the springs. However, the third material does not obey Hooke’s Law as the relationship between the force applied and the deformation of the spring is not directly proportional. From this experiment, it can also be concluded that Hooke’s Law only applies when a material is stretched within the elastic limit.

## References

BBC – GCSE Bitesize: Hooke’s Law, last updated/published in 【2014】, *BBC – GCSE Bitesize: Hooke’s Law* [ONLINE] **Available from: **

http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa/forces/forceselasticityrev2.shtml

Dynamics – Hooke’s Law Experiment, by JL Stanbrough, last updated/published on 【6 November 2002】, *Dynamics – Hooke’s Law Experiment *[ONLINE] **Available from: **

http://www.batesville.k12.in.us/physics/phynet/mechanics/newton3/Labs/SpringScale.html