course PHY 232
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Looks good, but there should be a succinct report of the quantities that varied and the quantities you obtained in your analysis (i.e., a couple of lines reporting tension and velocity so the reader can easily tell what tension went with each velocity in each string). You report two tensions for each but only one velocity, so I can't tell for sure what tension goes with each of the two velocities you report here.
String 1 (Twine String)Length of the string:
When attached to rubber bands the tension of the string stretched out the rubber bands.
With 2 Rubber Bands: 12.2 cm
with 3 "" "" "" "" "" : 10.2 cm
Using a spring scale to measure how much tension is in the rubber bands
With 2 Rubber Bands: 2.0N
with 3 "" "" "" "" "" : 1.75N
We used a pendulum in a certain set up to measure the time of the pulses in ""clicks"". The conversion from clicks to cycles is n clicks = [(1/2)n - (1/4)] cycles
We also know that the length of the pendulum is .098m so we can find the period of it by the expression 2pi/omega in which
omega= sqrt(g/L)
omega= sqrt[(9.8m/s²)/(.098m)]= sqrt[100 radians/s²]=10 radians/s
period = 2pi/omega = 2pi radians/(10 radians/s) = 2(pi)s/10 = pi/5 seconds = .628 seconds
So in the trial for string one, the pulse took 4 clicks to travel to one end and back to the beginning point. So it took [(1/2)4-(1/4] or 7/4 cycles to travel 72ft (twice the length of the string)
Using this we can calculate the time it took for it to travel 72ft and the average speed of the pulse. The time would be
period * cycles = pi/5 * 7/4 = 7pi/20 = 1.01 seconds
For the average speed of the pulse it would simply be 72ft/1.01s = 71.29 ft/s => 104.5mph
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Repeating the process for String 2:
String 1 (Nylon String)
Length of the string: 33 ft.
When attached to rubber bands the tension of the string stretched out the rubber bands.
With 2 Rubber Bands: 10.0 cm
with 3 "" "" "" "" "" : 9.6 cm
Using a spring scale to measure how much tension is in the rubber bands
With 2 Rubber Bands: 1.0N
with 3 "" "" "" "" "" : 1.1N
We used the same pendulum for the 2nd string as the first, so all period calculations are still valid into this trial.
So in the trial for string two, the pulse took 2 clicks to travel to one end and back to the beginning point. So it took [(1/2)2-(1/4] or 3/4 cycles to travel 66ft (twice the length of the string)
Using this we can calculate the time it took for it to travel 72ft and the average speed of the pulse. The time would be
period * cycles = pi/5 * 3/4 = 3pi/20 = .471 seconds
For the average speed of the pulse it would simply be 66ft/.471s = 140.13 ft/s => 205.52mph