Hypothesis
As the length increases, the distance that the mass has to move increases proportionally. This would suggest that the time for one oscillation would be directly proportional to the distance from the pivot; however, all the time the mass is falling it is accelerating so it would be less than proportional, and as it accelerates at a rate of km/s2, the time for one oscillation will be km/s2, and if the distance it must travel is directly proportional to the length we can replace m with l and get and get kL/ s2, and if s is the time for one oscillation, s can be replaced with p so we get kL/p2, and we can expand this to get p2 = kL, this can be written as p is proportional to the square root of l. Or P2 α L.
Results
For all of these readings, the mass was 200 grams.
All distance measurements are correct to the nearest two centimetres and the timings were correct to one hundredth of a second but my reaction time would have played some part.
There were two readings that were obviously very wrong which I omitted and repeated.
I measured the length of the string before and after to check that it had not slipped.
Conclusion
I predicted that the period would be proportional to the square root of the length of the string so to test this I drew a graph of the square root of the length against the period, hoping that I would get a straight line through the origin, and with this I would be able to work out what k is (in the formula p = k√L). I did get a straight line in which every point was within one per cent of the line. The gradient of this graph was 5.071, but to work out what k is, I must work out k = p/√L which is 1.972/10 = 0.197. Thus I have the formula p = 0.197*√L. This I will now check: for when L = 130, p = 0.197*√130
P = 2.25. I measured p in this case as 2.27, but this is still accurate to two significant figures.
Evaluation
For all of the experiment, the mass was kept the same (at 200 grams), the length of string was the only factor that I considered to be significant that was varied. I repeated each reading three times, and when I got an obvious rogue reading, I repeated it. In order to minimise the effects of my reaction time, I measured the time for five oscillations and then divided by five, as this decreased the influence of my reaction time five fold. The measurements of distance were to the nearest two centimetres and were measured from the centre of mass (I considered the string to be of negligible mass when finding the centre of mass, this may have been an oversight, but I could see no easy way of weighing it and keeping it straight out). The measurements of time were measured to the nearest hundredth of a second, but my reaction time would have made the last digit mostly irrelevant. However, I thought my reaction time would have been similar for each reading that it would add an extra bit of accuracy, and as the readings were generally so close, I left it in there.
As I said in the conclusion, every point was within one per cent of the straight-line p = 0.197*√130. This was very pleasing, and showed the accuracy of the experiment. Wind resistance obviously didn’t play a huge part, and the string was obviously thin enough not to impede the swing of the pendulum.
Resources used:
I looked up the formula on the internet site below, but as I understood it I put it in my hypothesis http://www.educationplanet.com/search/redirect?id=25782&mfcount=1&mfkw=pendulum&startval=0
