1: natural spring length (coils portion) = 18 cm Safe spring length = 54 cm 2:...

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1: natural spring length (coils portion) = 18 cm Safe spring length = 54 cm 2: mass of spring = 162 g Uncertainty of spring = +- 1 g 4: mass for safe stretch = 320 g 5 mass including weight holder = M 50 g 100 g 150 g 200 g 250 g 300 g 320 g 6: load near one fourth maximum Load including hanger = 80 g 50 oscillations, amplitude 2 cm t1= 18.95 sec. t2= 18.91 sec. T1=0.758. T2= 0.756 t3= 19.5 sec T3= 0.780 7:50 oscillations, amplitude 4 cm t1= 19.47 sec. t2=18.69 sec. t3= 18.98 sec T1=0.778. T2=0.747 T3= 0.759 8: Load near one half maximum Load including hanger = 160 g 50 oscillations, amplitude 2 cm t1= 24.37 sec. t2= 24.42 sec. T1=0.974. T2=0.976 t3= 24.28 sec T3= 0.970 9. 50 oscillations, amplitude 4 cm t1= 24.41 sec. t2= 24.23 sec. T1= 0.976. T2= 0.969 t3= 24.38 sec T3= 0.975 position 23 cm 29 cm 35 cm 41 cm 47 cm 52 cm 54 cm 1. Mean values of the period. 2. Theoretical values of the period. Best slope Maximum slope Minimum slope Load Of= m+m/3 Amplitude 2 cm 2 cm 4 cm 4 cm Amplitude 2 cm 2 cm 4 cm 4 cm T 0, T = 2π/w = 2√√ M Graphical Presentation Plot the stretching force in Newtons (vertical axis) against the position of the weight holder in me- ters. Remember to convert the added mass (in grams) to Newtons. Draw a straight line which fits best these experimental points. From the slope of the line determine a graphical average value of k. Pay close attention to the units of k. Carry out a maximum-minimum reasonable slope analysis to estimate the uncertainty in k, d Graphical Analysis T (9-7) 1: natural spring length (coils portion) = 18 cm Safe spring length = 54 cm 2: mass of spring = 162 g Uncertainty of spring = +- 1 g 4: mass for safe stretch = 320 g 5 mass including weight holder = M 50 g 100 g 150 g 200 g 250 g 300 g 320 g 6: load near one fourth maximum Load including hanger = 80 g 50 oscillations, amplitude 2 cm t1= 18.95 sec. t2= 18.91 sec. T1=0.758. T2= 0.756 t3= 19.5 sec T3= 0.780 7:50 oscillations, amplitude 4 cm t1= 19.47 sec. t2=18.69 sec. t3= 18.98 sec T1=0.778. T2=0.747 T3= 0.759 8: Load near one half maximum Load including hanger = 160 g 50 oscillations, amplitude 2 cm t1= 24.37 sec. t2= 24.42 sec. T1=0.974. T2=0.976 t3= 24.28 sec T3= 0.970 9. 50 oscillations, amplitude 4 cm t1= 24.41 sec. t2= 24.23 sec. T1= 0.976. T2= 0.969 t3= 24.38 sec T3= 0.975 position 23 cm 29 cm 35 cm 41 cm 47 cm 52 cm 54 cm 1. Mean values of the period. 2. Theoretical values of the period. Best slope Maximum slope Minimum slope Load Of= m+m/3 Amplitude 2 cm 2 cm 4 cm 4 cm Amplitude 2 cm 2 cm 4 cm 4 cm T 0, T = 2π/w = 2√√ M Graphical Presentation Plot the stretching force in Newtons (vertical axis) against the position of the weight holder in me- ters. Remember to convert the added mass (in grams) to Newtons. Draw a straight line which fits best these experimental points. From the slope of the line determine a graphical average value of k. Pay close attention to the units of k. Carry out a maximum-minimum reasonable slope analysis to estimate the uncertainty in k, d Graphical Analysis T (9-7)

Answer rating: 100% (QA)

To analyze the provided data and calculate the spring constant k for each set of oscillations we can use Hookes Law and the equation for the period of ... View the full answer