UB PHY 151LAB - Physics Lab M0
Pages 6

Unformatted text preview:

VI 1 Pendulum bob diameter measurement using ruler d 2 1cm d 2 4 cm d 2 59 cm Range of values for ruler is Maximum Minimum 2 59 cm 2 1cm 0 49cm 4 9mm Pendulum bob diameter using caliper d 24 8 mm d 24 8 mm d 29 5 mm Range of values for caliper is 29 5 mm 24 8 mm 4 7 mm 0 47 cm The range of measurements using the caliper was smaller than the range of measurements using the ruler I would therefore make the conclusion that the precision of the caliper is much more reliable than the precision of the ruler Thickness of the 100g mass using ruler t 0 9cm t 0 8cm t 0 8cm Range of values for ruler Maximum Minimum 0 9cm 0 8cm 0 1cm 1mm Thickness of the 100g mass using caliper t 9 2mm t 8 6mm t 8 5mm Range of values for caliper 9 2mm 8 5mm 0 7mm 0 07 cm Here the caliper was once again the more precise measurement than the ruler as it gave a smaller value So again in this case I would conclude that the caliper has a more precise measurement than the ruler The range of values for the ruler measurement of the pendulum bob is nearly 5 times larger than the caliper measurement of the 100g mass This is due to several factors The first is that the mass is narrower than the pendulum bob which gives a more accurate measurement This decreases the chance of experiencing parallax error The mass has flat edges compared to the bob which means you measure from the same spot each time Unlike the bob it is round and you would get different measurements each time you measure depending on where the measurement is taken from Average value of T given through excel 1 492 seconds Standard deviation of T as T given through excel 0 219 The standard deviation of the mean is given as T 0 0565 seconds T n 0 219 s 15 VI 2 Measured Values of T in seconds 1 11 1 29 1 47 1 66 1 66 1 07 1 66 1 56 1 59 1 63 1 17 1 62 1 6 1 72 1 57 Therefore T T 1 50 0 06 seconds VI 3 x0 xm cm 50 50 4 50 5 47 5 48 4 48 2 44 9 45 46 29 42 4 42 7 44 2 40 40 1 40 1 37 5 37 7 37 5 52 6 x cm 2 6 2 2 2 1 5 1 4 2 4 4 7 7 7 6 6 31 10 2 9 9 8 4 12 6 12 5 12 5 15 1 14 9 15 1 cm m g 20 20 20 40 40 40 60 60 60 80 80 80 100 100 100 120 120 120 Slope 7 69582634 4 41018227 Intercept Slope unc 0 21238458 2 03821959 Intercept unc x x0 xm When m 20 grams 52 6 cm 50 0 cm 2 6cm Mass vs Spring Elongation f x 7 7 x 4 41 0 2 4 6 8 10 12 14 16 Spring Elongation cm 140 120 100 80 60 40 20 0 s m a r g s s a M s s 7 70 0 21 g m b b 4 41 2 04 g Since m k g x the slope s k g Therefore k gs k 980cm s2 7 6958 g cm 7541 9 g s2 From equation 10 from the prologue with k C g A and s B we see that k k 7541 9 g Since g 0 k k s2 0 2124 g 7 6958 g cm k cm 208 1 g s2 k k 7542 208 g s2 VI 4 x0 xm cm 52 4 x cm 50 50 4 50 5 47 5 48 4 48 2 44 9 45 46 29 42 4 42 7 44 2 40 40 1 40 1 37 5 37 7 37 5 Slope 2 4 2 1 9 4 9 4 4 2 7 5 7 4 6 11 10 9 7 8 2 12 4 12 3 12 3 14 9 14 7 14 9 4 cm m g 20 20 20 40 40 40 60 60 60 80 80 80 100 100 100 120 120 120 7 6958263 5 94934754 Intercept 0 2123845 Slope unc 8 2 00059189 Intercept unc Values of x are determined the same way as in VI 3 The value of the slope calculated with the parallax error is precisely the same as was determined The calculation of k is carried out the exact same way as in VI 3 therefore The error analysis is also precisely the same as in VI 3 so by the calculation is VI 3 k 7541 9 g s2 k 208 1 g s2 k k 7542 208 g s2 The value of k is the same as determined in the calculation in VI 3

View Full Document

UB PHY 151LAB - Physics Lab M0

Documents in this Course

5 pages

5 pages

11 pages

3 pages

7 pages

13 pages

9 pages