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Drexel University *

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101

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Physics

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Dec 6, 2023

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12

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Simple Pendulum Lab Exercises PHYS 101 2020/2021 Complete the following exercises as a group and turn in a single document with your names. Anthony Moore Tyler Wilshire Sarenna Ben-Zeev _____________________ Show all work and calculations to receive full credit! You may use additional sheets. Note – each group member is to construct their own pendulum and make their own measurements, but all of you will report your results on the same worksheet and turn in one worksheet for the whole group. Answer the analysis questions as a team. Construct simple pendulums by following the directions in the lab manual. Each member of the group should have their own pendulum and take their own measurements. Paste in pictures of each pendulum in the space below.
Anthony Moore Sarenna Ben-Zeev
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Time Period and Length Measure the time period of your pendulum in the manner described in the lab manual. Make measurements for l = 20cm, 40cm, 60cm, 80cm, and 100cm by tying the string at different lengths. Each group member should make their own measurements. Put your data in the table below. T for 20cm T for 40cm T for 60cm T for 80cm T for 100cm TW AM SB Measurements from each group member 1.06s 1.33s 1.6s 1.76s 2.46s 1.15s 1.40s 1.80s 1.95s 2.38s 1.10 1.38 1.74 1.91 2.34 Using your T values, plot L vs T^2, as described in the lab manual. Add a trendline to your plot and take note of the slope in the trendline equation.
Tyler Wilshire Anthony Moore
Sarenna Ben-Zeev What is the slope of your trendline? (each member of the group should report their own value here) SLOPE = TW(0.0526)_ SLOPE = AM(0.0526)__ slope = SB(0.0515) As described in the lab manual, this slope is theoretically (4 pi^2 / g). So, you can use your slope to make an experimental prediction for g: g_exp = 4 pi^2 / (slope from trendline) Report your experimental predictions for g, g_exp, below. (each member of the group should report their own value here) g_exp = TW(750.54 cm/s^2) g_exp =AM(750.54 cm/s^2) __ g_exp = SB(766.57 cm/s^2) Now arrive at an agreed-upon group prediction for g_exp by finding the mean of every group member’s prediction. ( report just one value for the whole group here) <g_exp> = 755.88 cm/s^2or7.55m/s^2 Given that g = 9.8 m/s^2, how far off is your group prediction? Report a percent error using the
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procedure outlined in the lab manual. ( report just one value for the whole group here) Percent error in g_exp = 22.68%_ Knowing the uncertainties in the time and length measurements, it is possible to calculate the uncertainty in the experimental prediction for g, g_exp. See page 6 of the lab manual for details. This uncertainty will be different for each string length. Calculate this uncertainty for the l = 20cm and l = 100cm cases. Show your work and report your results below. Note: if members of the group used timekeeping instruments and length measuring instruments with different precisions, then you should report different values here. If the uncertainty is the same for all group members, then you should indicate that in your answer. Uncertainty for l = 20cm: 0.05s Uncertainty for l = 100cm: _0.06s Time Period and Amplitude Now investigate the relationship between amplitude, X, and period T. Keeping the length of your pendulums fixed at 100cm, measure the period in the same manner as before, while varying the amplitude. Measure T for X = 3cm, 10cm, 20cm, and 25cm. Again, each group member should make their own measurements, and the average of these will serve as the overall value for the group. Enter your data in the table below. **Time taken was only for one swing, not the average of 20 because my set up kept hitting things so it was impossible to take an average.(TW)** Average T: T for x= 3cm T for x= 10cm T for x = 20cm T for x = 25cm
TW AM SB Measurements from each group member 1.28 1.88s 1.953 2.03s 1.21s 1.76s 1.85s 1.98s 1.25 1.81 1.89 2.01 To determine if changing the amplitude had any effect on the period, plot x vs T and add a trendline with a trendline equation. Paste a picture of your plot below, or turn in your excel sheet with your worksheet. (each member of the group should put their own plots here) Tyler Wilshire
Anthony Moore Sarenna Ben-Zeev Following the procedure in the lab manual, calculate the relative percent variation in the T measurements from your x vs T data. Show your work and report your result in the space below. (each member of the group should report their own value here) Relative % variation = TW(36.94%) , AM(38.89%) , SB(37.81%) (δT /T) ×100 = (Tmax − Tmin / Tmax) ×100 TW- (δT /T) ×100= ((2.03-1.28)/2.01)*100
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(δT /T) ×100=36.94 AM - (δT /T) ×100 = (1.98 − 1.21 / 1.98) ×100 (δT /T) ×100 = 38.89 SB - (δT /T) ×100 = (2.01 − 1.25 / 2.01) ×100 (δT /T) ×100 = 37.81 Analysis Questions and Further Considerations: (answer these questions as a group) 1. How does your calculated uncertainty in g_exp for l=20cm and l=100cm compare to the percent error in g_exp? What does this tell you about how the length of the string affects the experimental prediction? The calculated uncertainty is slightly greater for l=100cm than l=20cm which in turn would mean that there is a greater room for error in the experimental prediction as the string gets longer. 2. How does the relative % variation in T for the x vs T measurements compare to the precision of your timekeeping instruments? If they are different, why might that be? To take time I used my phone, which is not a very precise method, which resulted in a large % variation. However, the percent variation was still too big for just imprecise timekeeping. The % variation is in large part due to bad time keeping but also the limitations of a cramped dorm may have also played a big part. 3. How would the period of a simple pendulum be affected if it were located on the moon
instead of the earth? ( g earth =6 g moon ). The period would dramatically increase in time as the gravity on the moon is 1.62ms^2. That being much less than the gravity on earth which is 9.8m/s^2. Since time can be found with the equation t=2pi(sqrt(l/g)), and the moon's gravity is smaller, the length divided by a smaller gravity and then squared will result in a larger value than that of earth's gravity. For example, on earth with a length of 5m it will take 4.49 seconds, but on the moon it will take 11.04 seconds over twice as long. 4. What effect would the temperature have on the time kept by a pendulum clock if the pendulum rod increases in length with an increase in temperature? If the rod expanded with an increased temperature, then the time of each swing would slightly increase due to an increased amplitude of oscillation. Over time this small difference would add up and the time kept would be behind the actual time. This can be explained with the equation for time t=2pi(sqrt(l/g)), increasing the l will increase the total time. This change in l will result in a very small time difference but over time the slight variation will result in the clock being way off. 5. We have neglected any effect due to air resistance on the motion of the pendulum. The justification for this is the assumption that the energy loss due to air resistance is a small fraction of the maximum kinetic energy of the pendulum. Suppose for the same fixed length of the string, you were to compare bobs made of steel, wood, and foam of the same size. How would the motion (the time period T and the amplitude of oscillation) be affected? The heavier the object, the less it is impacted by air resistance, A good example of this would be a feather and a bowling ball. With this knowledge, the Steel ball would be impacted very little, while foam would be impacted the most. The greater the impact, the longer the period will be. The steel ball would have the smallest T, followed by the wood, and lastly the foam block would take the longest and have the greatest T. The amplitude of oscillation would also be impacted, air resistance will cause a loss of energy. As a result each swing will result in a smaller amplitude of oscillation. The foam block is very light and so compared to the other two it has very little potential energy at the start, so losing energy from air resistance will impact it the
most. So the foams amplitude of oscillation will be reduced the fastest, followed by wood, and lastly steel. 6. When a mass attached to a spring is displaced by an amount x , the force acting on a mass is described by Hooke’s law, F = - kx where k is a constant known as the spring constant. This equation is similar to eqn. [3] in the manual. In the case of the mass-spring system, when displaced from equilibrium, the mass will undergo SHM with a time period. In the case of a simple pendulum, T does not depend on m. Both a spring-mass system and a simple pendulum follow similar force equations and undergo SHM. How would you explain, without using too much math, why T depends on the mass for a spring-mass system but not for the simple pendulum? In a spring mass system, T is dependent on mass because the mass will stretch out the spring by a distance of x. X is dependent on the mass, a heavier mass will stretch out the spring more. In the case of a simple pendulum the equation for T is t=2pi(sqrt(l/g)), mass is not a part of the calculation because, we assume that the pendulum is in a perfect vacuum, so all objects will have the same acceleration due to gravity and we can neglect m.
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