# Investigation of the relationship between extension of a spiral spring material per unit of load applied on it.

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Introduction

Investigation of the relationship between extension of a spiral spring material per unit of load applied on it.

Introduction

The objective of this investigation is to find out the relationship between the extension of a spiral spring per unit mass of load applied.

I believe that the following are two main factors or variables that affect in a big way the extension of a spring

Variables

- Firstly the force put on the spring which is the weights. I believe that the weights that are put on the spring will have the biggest affect on how the extents.

- Secondly the other variable is the extension. This is because the extension of the spiral spring changes with the load applied on it.

- Stiffness of the spring, an very stiff spring will extend lesser than a lesser stiff spring. This is because the particles are closer together so more load will need to be applied to weaken the bonds.

## Apparatus

The following equipment will be required in order to do this experiment:

- Firstly a clamp stand this will be used to this will be used to connect other apparatus in to one place. This will put the spring a good distanced up to prevent the spring from coming in to contact with the table which it will be on. The spring will be suspended from the stand to allow it enough space to stretch.

- Secondly two bosses will be used, one to hold the spiral spring in place and the other to hold the 1 meter ruler firmly in place. The bosses help secure all the apparatus in to place.

- Thirdly a weight hanger will be used to put the weights on the spring. The hanger will be connected to the bottom part of the spring.

- Also 100g weights will be put on the hanger one at a time.

- And lastly one last clamp will be used to clamp the bottom of the stand to the table. This is done to stop the stand from toppling over.

Middle

15

0.6

6

29.8

19

0.7

7

32.1

23

0.8

8

34.7

26

0.9

9

37.9

32

Mass on the hanger kg | Stretching force N | Scale reading /cm | Extension of the spring/mm |

0 | 0 | 23.6 | 0 |

0.1 | 1 | 24 | 4 |

0.2 | 2 | 2437 | 7 |

0.3 | 3 | 25.6 | 9 |

0.4 | 4 | 26.5 | 12 |

0.5 | 5 | 28.1 | 16 |

0.6 | 6 | 30 | 21 |

0.7 | 7 | 32.9 | 27 |

0.8 | 8 | 36.1 | 32 |

0.9 | 9 | 40 | 39 |

Mass on the hanger kg | Stretching force mg/n | Average Scale reading |

Conclusion

In the tables of results the stretching force was obtained my using the formula F=mg, where g = 9.8N/kg. The results show that my prediction at the start of the investigation was right, as the load applied on the spring increases, so does the extension of the spiral spring. So the two are proportion to each other, this is there relationship.

The graph that Is plotted on the pervious page is extension of the spring against the stretching force which is in newtons (N). when the line of best fit is drawn on the graph the plotted points come really close to it. This shows that the extension of the spring is directly proportional to the stretching force. In other words, if the stretching force is doubled the extension id doubled and so on. This relationship can be expressed as the following: Stretching force(F) Extension(X)

F=Kx

K= F/x

Now that the graph is completed the spring constant can be calculated by using the equation F=KX where K is the spring constant. The gradient(slope) of the line also gives the spring constant because stretching force divided by the extension. The formula for finding the gradient of a grade is as following:

= Y2 – Y1. = 9-0 = (0.253 + error)

X2 – X1 35.5-0

Spring constant (K) = (0.253 + error)

So the spring constant would be 0.253 in the force was in is in newtons, but if its in grams the answer would be( 25.49 + error).

This student written piece of work is one of many that can be found in our AS and A Level Waves & Cosmology section.

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