Safety:
As I am only timing the time period of oscillations of a spring there are only a few safety measures I need to take into account. Firstly as I am using weights I will place the retort stand no where near the edge of the workbench so if the spring breaks, it minimises the chance of the weights striking my foot or other such weak object near the floor. I will also wear safety goggles, as I will be placing strain on the metal spring and if it breaks its whiplash could be very damaging. I will make sure that the clamp holding the spring and weights are over the base of the retort so as to insure that the retort-stand will not fall over.
Theory:
In this investigation I shall be using the equation T2=4P2 x M
K
Where T is time period for one oscillation, M is mass used, and K is elastic constant.
Theory:
Oscillation is repeated motion back and forth past a central neutral position, or position of equilibrium. A single motion from one extreme position to the other and back, passing through the neutral position twice, is called a cycle. The number of cycles per second, or hertz (Hz), is known as the frequency of the oscillation.
A swinging pendulum eventually comes to rest if no further forces act upon it. The force that causes it to stop oscillating is called damping. Often the damping forces are frictional, but other damping forces, such as electrical or magnetic forces, might affect an oscillating spring.
I have predicted that if the spring is in a vertical line, then the amplitude of the mass will affect the time of 1 oscillation, because I think the mass will speed up as the amplitude goes up. That’s why the amplitude or the height will affect the time or the period of oscillations, but I will need to wait and see until I do the experiments to confirm this. I also said that the extension of the spring would be proportional to the load to a certain extent. This is called the elastic limit, where the spring’s strength doesn’t seem to follow its graph’s trend. Hooke’s law will relate to the extension, which is proportional to the load. Also the mass of the weight is inversely proportional to the frequency.
I said in my hypothesis that the amplitude does affect the time for one oscillation, and I will try to prove that. Here are the results, which were collected from a preliminary test.
As you can see from the preliminary data, most of the results for the time taken to do 10 oscillations are nearly the same, even though the amplitude was different. This means that the amplitude doesn’t matter when the load is dropped. So the height won’t affect the time of oscillations, it doesn’t matter where you let go of it, because it won’t affect the time of oscillations. So from now on, I will not measure the height (distance) anymore for each experiment.
The prediction that is stated above is supported by various theories and laws, which are explained below. First of all, it is common sense that the more springs used (series), the further down it will get pulled so the amplitude of the extension depends on the force. I have also stated the extension is proportional to the load, unless too much of a load is used which will cause the spring to go past its elastic limit.
Firstly, this prediction is from experiments with load and its effect on the length of 1 spring as it extends. I have found out that the extension was proportional to the load ‘E µ L’ to a certain extent. The certain extent is called the elastic limit. The beginning of the extension up to the elastic limit is called Hooke’s Law.
Once the elasticity limit of the spring has been passed, it loses its functions and ignores Hooke’s Law. In the beginning, the spring is very strong so the extension is very short, but as it passes the elastic limit, the spring’s extension gets longer and longer. Therefore I must be careful with this and must not put too much load onto the spring. It must be just the right and enough amount so as not to pass the elastic limit.
I believe the same pattern could be followed in my investigation, because I am using springs and the mass, where the current input variable in my experiment is the number of springs.
My prediction is made relating to the velocity in different parts of the oscillation which is based on Newton’s first and second laws. As I am investigating half an oscillation using a ticker-tape-timer, and I believe that at the 2 peaks, it will be at its slowest velocity and fastest at the centre of the oscillation (equilibrium).
The equilibrium state:
If Resultant force = Downward force
Then this is called ‘Equilibrium’
First of all by looking at the diagram, the resultant force is equal to the downward force and the mass and the spring remain unmoved due to Newton’s first law, and we call this equilibrium, where the 2 forces are equal. Now, when the spring is stretched by an external force, the weight moves down from its balanced position. There is now a displacement of the weight from its balanced position. it is greater than the downward force (gravity)/the upward force (pull of springs). Therefore, the spring and the weight accelerate upwards/downwards and the oscillation begins. The acceleration can be calculated with Newton’s Second Law
But in my case, the acceleration can be defined as this: If we used a larger mass, the same forces would be seen, but lower down.
it is common sense that as the number of springs increases, the lower the frequency (number of oscillation in a given time), and as the number of springs gets fewer, the higher the frequency.
The total amount of elastic potential energy and kinetic energy is constant; therefore there is always the transformation between the two forms of energy when the spring is oscillating. Therefore it has the basic and universal characteristics of all waves, which involves the transfer of energy.
Points to note for fair testing:
There are a few things that I will need to carry out in this experiment to keep it as fair as possible to get the most accurate results. I will use the same amount of load for every experiment. I will start my stopwatch as soon as the load comes up and is about to go down again. I will use springs, which have the same length to start off with. I will record the time to the nearest 2 decimal places.
Analysis:
As I said above I shall use two separate methods to work out elastic constant.
Both involve the above equation, T2=((4P2)/K)x M.
The calculation method involves rearranging that equation to give K as such
T2 = 4P2 x M K = 4P2 x M
K T2
The Graphical method means reading the first equation as a standard Y=mX+C straight-line equation.
T2 = 4P2 x M
K
Y = mX+C
Where Y= T2 , m=4P2/K and X=M, while C=0.
I can draw a graph with Time period squared (in seconds) on the Y-axis and mass (in kg) on the X-axis, and then plot my results. With that done I can then draw a line of best fit and work out its gradient using rise over tread.
Knowing the gradient (m) I can work out K as such:
m = 4P2
K
Which is the same as:
K= 4P2
m
So the elastic constant of the spring equals ((4P2)/ T2)x M and 4P2/gradient of plotted graph.
As K is equal to K = 4P2 x M
T2
We can say the units of K are equal to the units of M divided by the units of T2. So the units of K are kilograms divided by Seconds squared or rather Units of K = kgT-2
Results Table:
Analysis:
From the graph we can see that the gradient is equal to 0.5199 meaning that 4P2/K = 1.8104. Therefore K (the elastic constant) is equal to 4P2/1.8104 or rather 21.81 kgT-2
Using the calculation method I shall use the first third and fifth masses.
T2 = 4P2 x M K = 4P2 x M
K T2
1st reading:
K= 4P2 x 0.099
0.20
K = 19.54
2nd reading:
K= 4P2 x 0.296
0.56
K = 20.87
3rd reading:
K= 4P2 x 0.491
0.87
K = 22.28
As you can see the two recorded results are not that different. Using the Graph method I worked out an elastic constant of 21.81 kgT-2, and using the calculation method I got an average elastic constant of 20.90 kgT-2.
The results I got, complementing each other, do suggest a high degree of accuracy. From the graph I plotted we can see that there is a quantitative correlation between the mass on the spring and its time period for oscillations. This fits well with the theory that a constant, or rather the elastic constant, govern the time period of oscillations for a material. All the results I worked out fitted with the line of best-fit indicating that there are no anomalous results. Like wise the similarity of the values of K obtained from the calculation method indicates that there is a definite ink between mass and time period squared and this indeed is a co-efficient of the elastic constant. And the similarity between the results from both methods indicates that we have accurate results.
I need to find out the actual elastic constant of the spring I used to find out which, if either, is the most accurate method for working out the elastic constant of a material. Once that is found I can decide which method is best at working out the elastic constant.
Error analysis:
As in most experiments human error will have been the biggest offset. Things like judging when 10 cycles are finished and the reaction time in starting and stopping the timer will add to the inaccuracy of the experiment.
A large error I avoided was not taking the masses as what they were stated and measuring hem independently. As you can see above the masses were wildly different from their stated masses. At the end of this report you can see in a second graph how different the results would have been if I had used the stated masses. One error that I could not avoid was the time keeping accuracy. Not only was there human error (this was minimised by using the same human) but the stopwatch was only accurate to plus/minus 0.005 s. In a third graph you can see the potential error.
A large error was avoided by confirming the graph was plotted correctly. Without this correction I would have had an error of roughly 340% (Plotting the axis the wrong way round gives the elastic constant as roughly 75 and (75/22)*100 = 340).
If we drop the possible errors in to the equation we can work out the possible error in the elastic constant.
Avoided Mass error:
m = 1.78
K = 4P2/1.78 or rather 22.18 kgT-2
So by measuring the mass I avoided an error of 0.37, or, an error of 1.6%
Upper and lower range of Elastic constant:
Upper Gradient = 1.8243
Lower Gradient = 1.7965
Upper K = 4P2/1.82 or rather 21.69
Lower K = 4P2/1.79 or rather 22.05
So the elastic constant is 21.81 kgT-2 plus/minus 0.24,or, it has an error range of 1.16%
Having caught the potential error of mis-stated masses I’m quite confident of the experiments accuracy.
To improve this investigation I could rig up a more accurate timing device that used the oscillating load to start and stop the timer rather than just using a human observer. This would cut down on human error.
As an extension to this investigation I would study many different springs to confirm that the method applies to all materials, and was not just a chance fluke. I could also study whether heat and other environmental constants affect the elastic constant of a material.