An investigation into the effects of a force applied to a spring and the time for it to complete a set number of oscillations
An investigation into the effects of a force applied to a spring and the time for it to complete a set number of oscillations
Plan
Intro: I am going to investigate the effects of a force applied to a spring and its relationship to the time it takes for the spring to complete a set number of oscillations, when the displacement is constant. While conducting this investigation, I will always bear in mind, that I would like my results to be as accurate and reliable as possible. I have also previously conducted an investigation into the elasticity of a spring, which showed me that the force applied to a spring and its extension as a result of the force applied is directly proportional or constant. Could this imply that the frequency or the oscillations per second have a constant relationship to the force applied also?
Safety
Safety is paramount in all scientific investigations and this will be no exception. All masses and weights will be handled with care. Masses will not exceed the spring's maximum tolerance so there is no danger of the springs wire breaking. Food will not be consumed in the laboratory nor will drinks be drunk. People will not run in the laboratory. Safety spectacles will be worn so that if, by some unforeseen reason, the spring were to break, eyes will be kept safe.
Fair test
This will be a fair test by all other variables, within my control, being kept constant. The temperature for this investigation will be room temperature, approximately 23?C. The same spring will be used. The same displacement for the spring will be used. The same spring will be used because of differences in the spring constant. The number of oscillations the spring hast to complete will always be 10.
Scientific knowledge
When using springs in an investigation, it helps to revise/research a law called Hooke's Law. This is a formula that states that the force acting on a spring and the spring extension have a constant relationship. Which can be expressed by the simple equation:
F=KX
Force=constant x extension
Using this formula, we can find the constant of the spring if we know the extension and force acting on it by rearranging the equation
K=F/X
Constant=Force / Extension
and the formula can also be rearranged so that extension becomes the subject:
X=F/K
Extension =Force / Constant
F= Force
K= Constant
X= Extension
I believe that momentum is also worth mentioning here because I think that momentum is what will have the time delaying effect on the spring's oscillation time. Momentum is an object acting under its own kinetic energy that has stored. If an object was to be accelerated, kinetic energy being transferred into the object, and then to stop accelerating and for external forces to stop acting on it, the object can continue to move under its own kinetic energy until the energy is transferred into other objects or particles, in the case of air resistance. Momentum can be expressed by the simple formula:
Momentum = mass x velocity
Kg per m/s2 = kg x m/s
The force that the spring can exert is always constant (unless the spring elastic limit has been reached), this is important to understand first to understand why the spring oscillation time increases when more mass is applied to the spring.
When the mass is displaced by 10cm, energy form the person that does the displacing is transferred into potential energy in the spring. So, when the spring is released, the spring 'uses' this potential energy to bring the mass back to its resting position. But as the mass is moved up to the rest position is ...
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The force that the spring can exert is always constant (unless the spring elastic limit has been reached), this is important to understand first to understand why the spring oscillation time increases when more mass is applied to the spring.
When the mass is displaced by 10cm, energy form the person that does the displacing is transferred into potential energy in the spring. So, when the spring is released, the spring 'uses' this potential energy to bring the mass back to its resting position. But as the mass is moved up to the rest position is gains momentum, so when it reaches the rest position the momentum of the mass keep it moving up wards. This 'upward' momentum keeps going until the momentum and gravity forces are balanced (the momentum is converted into gravitational potential energy at the point it passes the rest position) at which point the mass starts to move down under the influence of gravity. Once the mass has passed the rest point the GPE is converted in to potential energy in the spring, this continues until the forces are balanced and the cycle stars again. All the time energy is being lost as friction and in air resistance, so eventually the mass does come to rest at the rest position.
This is a fairly general explanation of way the spring oscillates but not of why the oscillation speed decreases with more mass. To explain this we must use Newton's second law.
F=ma
N=kgxm/s2
If the force remains constant then the mass and the acceleration are variable. If the mass was to change but the force to remains proportionately the same then the acceleration would change to compensate (the force that the spring exerts is equal to the force that is displacing it).
E.g.
F=ma
N=0.1kgxa
N/0.01kg=a
a=100m/s2
That is quite a fast oscillating spring. But if more mass was added to the spring:
F=ma
5N=0.5kgxa
5N/0.5kg=a
a=10m/s2
(Notice that in the second example the force that the spring exerts changes. This is because the force acting on the spring has changed from 1N to 5N and the force that spring can exert is the same as the force acting on it.)
Which is much slower.
So, I have explained why the spring oscillates and why the increase in mass increases the oscillation time. I feel I have sufficient knowledge to conduct this investigation.
If the movement of the mass was plotted, in real time, on to paper against the time the 'oscillation' graph could be said to display properties much like that of a wave. If the movement of the mass were plotted against time, a wave would be visible. See attached paper. As you can see the relationship between the movement of the mass and time is very wave like. Now working on the basis that oscillation is a wave several things can be found: the amplitude (from the point of rest), the wavelength (time to complete one wave), and most importantly the frequency. Now frequency is the most important thing to find on a wave graph in an investigation to find the effects of mass/force on the rate of oscillation because it tells us the number of oscillations per second, which would be very useful in establishing which certain force/mass acting on a spring is faster to completer 10 oscillations or not compared to anther. This knowledge, however, has little relevance to this particular investigation into spring oscillation because the oscillations can simply be timed, but it is worth noting.
Prediction
I predict that as the force acting on the spring increases, so will the times for it to complete the 10 oscillations because as proven in the previous example the mass on the spring will affect the acceleration which will directly affect the frequency (oscillations per second).
Diagram
Equipment
The force acting on the spring is the force that the masses can exert (their weight) where 100 grams equal to 1 Newton. For the mass to effectively transfer to the spring the mass should be vertically in line with the spring. The spring is manufactured with a hook on both ends. Which means that I can suspend the spring from a Clamp and with the aid of a piece of apparatus with a hook designed on one end and the mass of 100 grams I can attach this to the other hook and add masses to this arrangement. Thus, I have placed the masses vertically in line with the spring. Because I've suspended the spring from the clamp, I'll also need a retort stand and a boss. To make this set-up safe I will place a counterweight at the base of the recall stand with a mass of 5 kilograms, so that he will be greatly heavier then anything I will place on the other side of the set-up. I will use a 1-meter ruler to measure displacement from the base of the mass hook (10cm). I will also use a stopwatch departing how it takes for the spring complete 10 oscillations. For safety reasons selo-tape will be placed around the joint of the spring and the mass hook, so that if the acceleration of the spring is too great and mass hook were to gain enough momentum to come off the hook it would be unable to do so.
Equipment that I will need in this investigation include:
* 1 Retort stand
* 1 Boss
* 1 Clamp
* 1 One meter ruler
* 1 Spring
* Masses
* Counter weights
* 1 one meter ruler
* Selo-tape
Method
The range for this investigation will be 10 Newton's. And will be repeated 10 times to gain accuracy and reliability. I have produced a step-by-step method plan of what I'm going to do. Safety will be in mind at all times during this practical. Measuring with the ruler will be done to the best of my ability to make sure that the results are as accurate and as reliable as possible. The timing will be done to the best of my ability for the same reasons as the measuring.
. Set up the apparatus as shown in the diagram.
2. Add 100g to the spring and use hands to lower the mass to the resting position.
3. Displace the spring by 10cm downwards.
4. Let go of the spring and time how long it takes in seconds to complete 10 oscillations
5. Record this time on the table.
6. Add 100g to the spring and use hands to lower the mass to the resting position.
7. Displace the spring by 10cm downwards.
8. Let go of the spring and time how long it takes in seconds to complete 10 oscillations
9. Record this time on the table.
0. Add 100g to the spring and use hands to lower the mass to the resting position.
1. Displace the spring by 10cm downwards.
2. Let go of the spring and time how long it takes in seconds to complete 10 oscillations
3. Record this time on the table.
4. Add 100g to the spring and use hands to lower the mass to the resting position.
5. Displace the spring by 10cm downwards.
6. Let go of the spring and time how long it takes in seconds to complete 10 oscillations
7. Record this time on the table.
8. Add 100g to the spring and use hands to lower the mass to the resting position.
9. Displace the spring by 10cm downwards.
20. Let go of the spring and time how long it takes in seconds to complete 10 oscillations
21. Record this time on the table.
22. Add 100g to the spring and use hands to lower the mass to the resting position.
23. Displace the spring by 10cm downwards.
24. Let go of the spring and time how long it takes in seconds to complete 10 oscillations
25. Record this time on the table.
26. Add 100g to the spring and use hands to lower the mass to the resting position.
27. Displace the spring by 10cm downwards.
28. Let go of the spring and time how long it takes in seconds to complete 10 oscillations
29. Record this time on the table.
30. Add 100g to the spring and use hands to lower the mass to the resting position.
31. Displace the spring by 10cm downwards.
32. Let go of the spring and time how long it takes in seconds to complete 10 oscillations
33. Record this time on the table.
34. Add 100g to the spring and use hands to lower the mass to the resting position.
35. Displace the spring by 10cm downwards.
36. Let go of the spring and time how long it takes in seconds to complete 10 oscillations
37. Record this time on the table.
Observation
After having conducted the practical safely and as accurately as can, I have, using the results, drawn up this table. This is the results table for the 10 experiments:
Results for practical 1,2,3 etc
Time it took the spring to complete
0 oscillations in seconds
Force acting on spring (N)
2
3
4
5
6
7
8
9
0
3.75
4.00
3.81
3.63
3.75
3.56
3.53
3.60
3.70
3.43
2
5.50
5.34
5.81
5.50
5.47
5.34
5.40
5.57
5.47
5.48
3
6.32
6.38
6.28
6.44
6.50
6.53
6.66
6.53
6.56
6.44
4
7.59
7.31
7.25
7.59
7.40
7.47
7.66
7.50
7.56
7.68
5
8.37
8.03
8.06
8.38
8.41
8.41
8.43
8.40
8.34
8.44
6
9.19
8.84
9.00
9.50
9.03
9.13
9.12
9.25
9.03
9.09
7
0.00
9.72
9.78
9.72
9.78
9.81
9.70
9.78
9.87
9.72
8
0.50
0.22
0.10
0.38
0.34
0.38
0. 34
0.38
0.37
0.35
9
1.16
0.62
0.62
0.87
0.87
0.98
0.94
1.00
0.56
1.06
I originally planed to go to 10N but this broke the spring's elastic limit and spring would no longer retake its original shape and the results became erratic, so the information was not included in the results table.
From my results I have plotted a graph of time for spring to complete 10 oscillations against the force applied to the spring to see if there is a relationship between the between the two factors.
Analysis
The graph shows that the relationship between the force applied to the spring and the time it took to complete 10 oscillations is not constant. They have a non-uniform relationship. I have also included the rate of oscillation, that's, rate of oscillation per second (found by 1/time= rate/sec). The range between the times seems to decrease as the mass increases, from a range of 1.812 at 1N to a range of 0.5325 at 9N. This change in, range differences, gives a curved line of best fit. Also the range decreases more so as the first 3N's of force are applied, then after.
Force acting of spring (N)
Average Time in seconds
Rate of oscillation
3.676
0.272034820
2
5.488
0.182215743
3
6.464
0.154702970
4
7.501
0.133315557
5
8.327
0.120091269
6
9.118
0.109673173
7
9.788
0.102165917
8
0.336
0.096749226
9
0.868
0.092013249
There appears to be no anomalous results. The graph also shows an unexpected relationship, the curved gradient, this could be explored more in a different investigation. It probably has some thing to do with the spring approaching its elastic limit. The table quite clearly shows that as more force acts on the spring, it takes longer to complete the set 10 oscillations the rate decreases. This is what I predicted. The curved gradient is some thing that I never even considered and could either be the result of me not taking in to account some other variable or that there was errors in the measuring of the variables or maybe this is complete relationship all by it self.
Evaluation
I think that the set up and equipment used in this investigation was correct and proper for the task at hand. I feel, although not too shore because of the result obtained, that the accuracy was good if not a little excessive in places e.g. the rate of oscillation. I am currently not shore on any improvements for the procedure but if where to conduct a more detailed investigation I would invest some time into finding a more accurate or even reliable method. I believe that the results are reliable to draw up simple conclusions on the behaviour of springs under different forces but I would prefer a second opinion before I say, with any real faith, any more complexes theories on this investigation. I would like to do this investigation again but I would have the variable as the displacement to see if that has any other effects on the rate of oscillation. It could alter the rate by increasing the amplitude of the oscillation, which if the acceleration of the mass were constraint, would take the mass longer to complete the 10 oscillations. I feel it would be some thing worth investigating, may it help me understand what is happening in this investigation more.
Keith Thompson
