PHY158 Lab Report M0

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Dec 6, 2023

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VI-1 Pendulum Bob Diameter, d: Ruler measurements (cm) 2.40 2.30 2.40 Caliper measurements (mm) 24.9 25.0 25.1 Range for ruler measurements is maximum − minimum = 2.40 cm − 2.30 cm = 0.10 cm or 1.00 mm Range for caliper measurements is maximum – minimum = 25.1 mm − 24.9 mm = 0.20 mm or 0.02 cm The equations above show that the Caliper has a smaller range indicating that it is more precise than the ruler. 100 g mass thickness: Ruler measurements (cm) 0.80 0.60 0.65 Caliper measurements (mm) 8.50 8.70 8.70 Range for ruler measurements is maximum – minimum = 0.80 cm – 0.60 cm = 0.20 cm or 2.00 mm Range for caliper measurements is maximum – minimum = 8.70 mm – 8.50 mm = 0.20 mm or 0.02 cm The equations above show that the Caliper has a smaller range meaning it is more precise. The size of the range of ruler measurements of the pendulum bob is smaller than the size of the range of ruler measurements of the 100 g mass. Considering that the 100 g mass has a flat surface and isn’t a sphere, it is unusual that the range for the pendulum bob would be smaller. There is one measurement for the 100 g mass that is an outlier, 0.80 cm, which leads me to believe there was human error when measuring the 100 g mass. VI-2 Measured Value of T (sec) Student 1 Student 2 1.78 1.81 1.88 1.50 1.71 1.60 1.90 1.78 1.80 1.51 Excel returns the average value of T = 1.727 seconds Excel returns the standard deviation of T as σ T = 0.14384 seconds The standard deviation of the mean us given as σ ´ T = σ T √ n = 0.14384 √ 10 = 0.045486 seconds Therefore T ±σ ´ T = 1.73 ± 0.045 seconds

VI-3 X 0 = 59.2 cm X = X 0 – X m Ex.) m = 80 g, 59.2 cm – 49.0 cm = 10.2 cm m (g) X m (cm) X (cm) 20 56.8 2.4 20 56.9 2.3 20 56.8 2.4 40 54.2 5.0 40 54.1 5.1 40 54.2 5.0 60 51.7 7.5 60 51.6 7.6 60 51.5 7.7 80 49.0 10.2 80 49.1 10.1 80 49.1 10.1 100 46.2 13.0 100 46.3 12.9 100 46.2 13.0 120 43.8 15.4 120 43.9 15.3 120 43.8 15.4 Slope 7.660955 1.732382 Intercept Uncertainty of Slope 0.039623 0.394799 Uncertainty of intercept

0 2 4 6 8 10 12 14 16 18 0 20 40 60 80 100 120 140 f(x) = 7.66 x + 0.97 m vs x spring Elongation, cm Mass, grams Since m = kx g and s = k g gives us k = gs k = ( 980 cm s 2 ) ( 7.660955 g cm ) = 7507.7359 g s 2 Eq. P-9: k = f, g = x, and s = y gσ s ¿ ¿ ( sσ g ) 2 + ¿ σ k = √ ( ∂k ∂g σ g ) 2 + ( ∂k ∂s σ s ) 2 = √ ¿ σ g = 0 σ k = gσ s = ( 980 cm s 2 ) ( 0.039623 g cm ) = 38.8 g s 2 k ±σ k = 7508 ± 40 g / s 2

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VI-4 X 0 = 59.3 cm Values of X are determined the same way as in VI-3 Slope and Intercept (including their uncertainties) were calculated the same way as in VI-3 Using the equation for K in VI-3: k = ( 980 cm s 2 ) ( 7.660955 g cm ) = 7507.7359 g s 2 which is the same answer as in VI-3 m (g) X m (cm) X (cm) 20 56.8 2.5 20 56.9 2.4 20 56.8 2.5 40 54.2 5.1 40 54.1 5.2 40 54.2 5.1 60 51.7 7.6 60 51.6 7.7 60 51.5 7.8 80 49.0 10.3 80 49.1 10.2 80 49.1 10.2 100 46.2 13.1 100 46.3 13.0 100 46.2 13.1 120 43.8 15.5 120 43.9 15.4 120 43.8 15.5 Slope 7.660955 0.966287 Intercept Uncertainty of Slope 0.039623 0.398347 Uncertainty of intercept

Using the equation for σ k in VI-3: σ k = ( 980 cm s 2 ) ( 0.039623 g cm ) = 38.83054 g s 2 which is the same answer as in VI-3 The value of k is the same as determined in VI-3 k ±σ k = 7508 ± 40 g / s 2

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