m g xm cm 1 2 3 4 5 6 37 6 35 32 5 29 9 27 3 24 7 20 40 60 80 100 120 x x0 xm cm 2 5 5 1 7 6 10 2 12 8 15 4 m vs x f x 7 76 x 0 68 VI 1 Holder mass mh 5 1 g Spring mass ms 2 8 g Flag mass mf 3 5 g x0 40 1cm x 0 0cm m c m 140 120 100 80 60 40 20 0 0 2 4 6 8 10 12 14 16 18 Using LINEST function in Excel the following values were obtained x cm We can compare the force on a spring with the slope intercept form of the line that was calculated using LINEST in Excel Thus we can then determine the spring constant k Slope s 7 76025237 y intercept b 0 67507886 Uncertainty in the slope s 0 02329807 Uncertainty in the y intercept b 0 23201965 Calculating the spring constant We use the equation y sx b y sx b s kg k s g k 7 76025237 g cm k 7605 05 g s2 Calculating k k k 661 5773 g s2 K k k 7605 6 62 10 2 g s2 VI 2 m g mT g 48 6 40 C1 240 C2 234 C3 239 68 6 88 6 203 203 200 181 178 179 108 6 161 149 161 128 6 149 148 149 N1 119 75 101 25 90 25 80 25 70 24 N2 116 75 101 25 88 75 74 25 73 75 N3 119 25 99 75 89 25 80 25 74 25 T1 s 0 5010 4384 0 5925 9259 0 6648 1994 0 7476 6355 0 8542 1412 T2 s 0 5139 1863 0 5925 9259 0 6760 5634 0 8080 8081 0 8135 5932 T3 s 0 50314 2 s 2 T1 0 251044 2 s 2 T2 0 26411 2 s 2 T3 0 25315 0 60150 0 351165 0 35116 0 36180 0 67226 0 441985 0 45705 0 45194 0 74766 0 559000 0 65299 0 55900 0 80808 0 729681 0 66187 0 65299 465 376 891 355 081 454 677 548 079 459 236 598 217 459 877 93 98 56 79 77 60 80 100 120 4 4 To determine the values of N T and T2 2 Ci 1 1 N T 60 00 N 2 T 2 T i Calculations for Trial 1 are on next page N 2 240 1 1 119 75 479 4 T 60 00 N 60 00 119 75 5010438143 T 2 T i 2 50104381432 0 2510449309 T2 vs m f x 0 01 x 0 s 2 T 0 8 0 7 0 6 0 5 0 4 0 3 0 2 0 1 0 30 Using LINEST function in Excel the following values were obtained Slope s 0 00582554 y intercept b 0 00053241 Uncertainty in the slope s 0 00044694 Uncertainty in the y intercept b 0 03792371 40 50 60 70 90 100 110 120 130 80 m g Comparing to slope equation y sx b we can further assume that s 4 2 k Calculating k and k k T 2 m ms We square both sides to then get T 2 4 2 m ms k T 2 4 2 m 4 2 ms k k 4 2 k 0 00582554 s2 g 6776 7825 g s2 Solving for k we get k 4 2 s s 2 s 2 k k s 2 k 4 2 s2 k k 3 029 K k k 6777 3 03 101 g s2 Both methods that were calculated in VI 1 and VI 2 do not yield the same value of k although they are somewhat close to one another VI 3 We can compare this with the slope equation y sx b b 4 2 ms T 2 4 2 k 4 ms k k bk 4 2 ms 0 00053241 0 03515 VI 4 The range of values indicated within the introduction of the lab in the manual does not match the amount provided in this step The inaccuracy could be caused by numerous things Looking at my graph we can see that the third and fifth points differ the most compared to the other data points This would result in the greater value of and a higher y intercept value d 2 6 cm d 0 cm l l l d 2 C1 C2 C3 N1 N2 N3 T1 s T2 s T3 s T1 2 s 2 T2 2 s 2 T3 2 s 2 30 3 31 6 105 106 106 52 25 52 75 52 75 1 14832536 1 13744076 1 13744076 1 31865113 1 29377148 1 29377148 34 8 40 47 8 57 5 65 5 36 1 41 3 49 1 58 8 66 8 93 93 84 79 74 99 93 83 80 75 99 93 85 75 72 46 25 49 25 49 25 1 2972973 1 21827411 1 21827411 1 68298028 1 48419181 1 48419181 46 25 46 25 46 25 1 2972973 1 2972973 1 2972973 1 68298028 1 68298028 1 68298028 41 75 41 25 42 25 1 43712575 1 45454545 1 42011834 2 06533042 2 11570248 2 01673611 39 25 39 75 37 25 1 52866242 1 50943396 1 61073826 2 3368088 2 27839089 2 59447773 36 75 37 25 35 75 1 63265306 1 61073826 1 67832168 2 66555602 2 59447773 2 81676366 l vs T2 f x 0 04 x 0 26 2 s 2 T 2 5 3 2 1 1 5 0 5 0 25 Using LINEST function in Excel the following values were obtained Slope s 0 03585164 y intercept b 0 26353267 Uncertainty in the slope s 0 00277537 Uncertainty in the y intercept b 0 1356717 The equation of a pendulum is given by T 2 l Squaring both sides we get T 2 4 2 g l g 30 35 40 45 50 55 60 65 70 L cm Comparing this with the slope equation y sx b we get the following Solving for g we get the following 0 03585164 1101 161 cm s2 s 4 2 g 4 2 g 4 2 s g Solving for g g 4 2 g 2 s2 g 85 24 cm s 4 2 0 03585164 2 2 g g 1101 85cm s2 The value g 980 cm s2 does not fall within the range defined by my value and its uncertainty As a matter of fact it is not even remotely close to the value of 980 cm s2 which could signify that there was some human error when recording data throughout the experiment