An Investigation into the Oscillation of a Spring.

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An Investigation into the Oscillation of a Spring

Introduction

When you have a spring suspended in mid air with a weight on it, unless you supply it with energy or gravitational potential energy it will stay still.  However when supplied with energy, whether its elastic potential energy or gravitational energy it will oscillate back and forth from the point of rest.  With this in mind and also keeping in mind the below equations and rules we are looking to address the matter of the speed of the oscillation of the spring under different weights.

T=2*pi* mass/k

Where K is a constant that occurs in all springs but changes with each spring to show the relationship between mass and time.

Force=mass*acceleration

Therefore a smaller mass would equal a larger acceleration.

Hookes law

Stress=modulus*strain

Where stress is the force applied per unit area and strain is relative to any change taking place when stress is applied. And modulus is a constant describing the relationship between stress and strain

Aim

To investigate how the mass of the weights affect the speed of the oscillations

Method

We will set up the apparatus as shown in the diagram below.

The first thing we did, after setting up the equipment, was to set the spring to its rest point as this was essential for a fair test. We would then pull the spring down a certain length we chose two centimetres.  Then as one oscillation would be too fast to measure we would time ten oscillations from being pulled down two centimetre. After we have done this three times with a weight hanging off it, say 100g, we would then add another 100g to the spring and repeat the same process, making sure that the distance we pulled the spring down by was equal to that of before, e.g. two centimetres.

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Hypothesis

I believe that the more weight you put on the spring, the more time it will take to make an oscillation. To explain this I will start off at the beginning. According to Hooke’s Law the extension is directly proportional to the force loaded onto it. The graph below shows the load and the extension. The line is straight meaning x=y or extension is directly proportional to load.

And therefore that if you doubled the weight you added to the spring then the extension of the spring will double as well.

However this ...

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