# Investigation of Ellasticity

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[pic 1]

INTRODUCTION:

An elastic body is the one that will return to its original shape. In this case, elasticity of a spring is going to be focused upon. Elasticity is the tendency of a body to return to its original shape after it is stretched or compressed.

The more elastic a spring is, the easier it is to deform. When a spring is compressed or stretched, the force it exerts is proportional to the change in its length. According to Hooke’s law (www.britannia.com, 2012) the force F applied on a spring is directly proportional to the extension x of the spring: [pic 2]

[pic 3]

Designate x by F = -k.[pic 4][pic 5]

The magnitude of the value of k depends upon different types of springs. k is larger for the springs which are harder to pull and to stretch and is small for those that can be easily stretched.

If the applied forces do not exceed the proportional limit, Hooke’s law can be applied. If the spring is stretched too far where it exceeds its elastic limit, it does not return back to its original shape.

The aim of this experiment is to determine the spring constant of the three springs with different elasticity.

Since F applied to a spring is directly proportional to the extension ∆l, it can be hypothesised that the value of the spring constant is large for stiffer strings whereas smaller for easily stretchable springs. [pic 6]

[pic 7]

MATERIALS AND METHOD:

Materials:

• Ruler
• Mass hanger (mass of 10g)
• Steel springs (3)
• Retort stand
• 10g weights

[pic 8]

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Figure 2: The apparatus set up to measure the extension of the spring

Method:

1. Firstly, the apparatus was setup as shown in figure 2.
2. Next, the length of the first massless spring was measured using a ruler.
3. Later, the mass hanger was hung to the spring and the extension was measured which was 10g. Then loads of masses ranging from 10g to 100g were hung to the end of the spring (figure 2) and the length l2 was measured after adding each load.
4. After that, extension ∆l was calculated for each mass.
5. Steps 2- 4 were repeated using two other springs

RESULTS:

Table 1: Extension calculated for spring 1

 M – Mass (g) l2  (mm) ∆l = l2 – l1 (mm) 0 20 0 10 20 0 20 20 0 30 20 0 40 21 1 50 22 2 60 23 3 70 24 4 80 25 5 90 27 7 100 28 8

Table 2: Extension calculated for spring 2

 M – Mass (g) l2  (mm) ∆l = l2 – l1 (mm) 0 20 0 10 20 0 20 20 0 30 20 0 40 21 1 50 24 4 60 27 7 70 31 11 80 34 14 90 38 18 100 41 21

Table 3: Extension calculated for spring 3

 M – Mass (g) l2  (mm) ∆l = l2 – l1 (mm) 0 20 0 10 20 0 20 21 1 30 24 4 40 27 7 50 31 11 60 35 15 70 38 18 80 42 22 90 45 25 100 49 29

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