So, when the spring combinations are set up, the mass will be pulled down so that it oscillates. 10 oscillations will be measured by a using a stopwatch. I am measuring 10 oscillations instead of one because then the time will be more accurate as it is very difficult to measure one oscillation because it can be very fast and human reaction time could make the time very inaccurate, because the stop watch will be pressed at the wrong times. So by counting 10 there is less of margin of error. Then to get the result for one oscillation the time will simply be divided by 10. This process will be repeated for all the different spring combinations. The results will be repeated three times so then the average can be taken, this way it will make the final result even more accurate and it will eradicate any anomalies.
The mass used will be six, 50g weights. So in total this is 350g. The different spring combinations that will be used are…
- 3 parallel
- 2 parallel
- 2 parallel and 1 series
- 2 series
- 3 series
- 4 series
PRELIMINARY EXPERIMENT
A preliminary experiment was carried out to find out the spring constant (stiffness). The 50g masses were added on one by one, and the extension was measured by a ruler. This was also carried out for each different spring combination, to work out each one. Then the results collected, and the spring constants were worked out.
The stiffness for a single spring is 21, so then you can see the spring stiffness for the other combinations in relation. For example, for 2 in series, the spring constant is half of a single spring, 3 in series, the spring is a third of that of a single spring, 4 in series, and it is a quarter. For 2 in parallel, the spring constant is double that of a single spring, for 3 in parallel it is triple. For 2 in parallel and one series, the spring constant is 2/3 of a single spring. This proves what I said in the planning.
I predict that the stiffer the spring will be, the shorter the oscillation period will be also.
The equation that connects spring stiffness and oscillation period is:
We can ignore 2╥ because it is constant, so therefore what we are left with means that the period is inverse proportional to the square root of the spring stiffness. The formula then, to work out the period can look like the following:
T= 1
√ K
This formula shows that the period and the spring constant are in fact inverse proportional because it shows as one figure increases the other one would decrease. This is shown by K being a denominator in the formula.
ANALYSE
As you can see from the graph, the line suggests that there is an inverse relationship between the period and the spring constant, although they are not in direct proportionality to each other. The formula on the graph shows that period is equal to 3.6736 times the spring constant, more or less to the power of a negative half. I know that to the power of a half is the same as the square root, and negative power means that number divide by 1. So altogether this means that the periods is equal to 3,8425 times by 1, and divide by the square root of the spring constant.
T = 3.825
√k
The number 3.6736 came from the formula 2╥√M, so in this case the mass used in the experiment was around 350g, or 0.35 Kg. The square root of this is 0.59. So then 0.59 multiplied by 2╥ is 3.7172. This is more or less the number I got from the results all thought it is a bit off. It’s not exact, and this is due to the fact there probably would have been a small margin of error because of human reaction time in the timing.
Conclusion
The oscillation period decreases as the spring gets stiffer. This is because if two springs are in parallel then the there is twice the stiffness so then the mass is shared, and the force is halved. The period is reduced because the spring extension is shorter and the oscillations are completed faster.
From the results I can deduce that the oscillation period is in inverse proportionality to the square root of the spring constant.
EVALUATION
I would say that the results of the experiment are actually quite reliable. I think they are reliable enough to support the conclusion as the results gave me a graph that showed that oscillation period is in inverse proportionality to the square root of the spring constant. The graph did produce the right curve however some of the points weren’t exactly on the line. However, they were accurate enough to give me the right results and the formula on the graph was accurate enough to work.
The results were quite good because if you look at the table the results are all very similar, so this shows the timing was quite good. Also, averages were taken, so this gave a more accurate figure. The figures used were all to two decimal places so the figures used were actually quite precise.
The errors will most probably have been a cause of human reaction time. Humans can’t time the oscillations so precise because they’re reactions aren’t perfect. This is why some results may have been slightly out of place. However, they weren’t that out of place that it affected the investigation in any major way. The only other way, human error could be eradicated, is by using some sort of mechanical way instead.
Other work that could be done is to investigate how changing the masses effects the oscillation period. The same apparatus could be used too and the same formula would apply.