Fluent Essays

Introduction to Uncertainty What is experimental uncertainty? In experimental work, we generally assume there is a “true” value for the quantity we measure if we have a perfect experimental design, and perfect instruments in a perfect laboratory. However, thanks to various imperfections in real life, real measurements always turn out to be different from the “true” value to a certain degree. We say that the measurement results carry uncertainty. Is experimental uncertainty the same as human error? Experimental uncertainty is NOT human error. Experimental uncertainty is caused by inevitable factors, for example limited instrument precision, experimental design limited by budget or technology, etc. We can tweak our experimental method to minimize uncertainty, but there is no way to completely eliminate it. Therefore, it is not a bad thing to report experiment results with uncertainties; in fact, it is ethically right to report experimental results with uncertainties. On the other hand, human error is a gross mistake in experiment design or operation, it is avoidable, and should be eliminated. We hope this explanation can help you distinguish the difference between the two terms and use them correctly from now on. Throughout the physics labs this distinction will be continuously assessed. How to do uncertainty analysis Since we do not know the “true” value, (otherwise, we do not need to measure it in the first place) what can we do to find out how reliable our measurement is? We need to resort to uncertainty analysis. There are many different ways for uncertainty estimation, we will learn one method in this class. Mean (average), standard deviation, standard error of the mean and confidence level When we repeat the same measurement multiple times, we usually will not obtain the same reading every single time. In the first experiment, we measure the diameter of a metal rod multiple times, let us use it as an example. The table below summarizes a total number of counts 𝑁 = 10 reading results. Reading count 𝑛 1 2 3 4 5 6 7 8 9 10 Diameter (mm) 6.012 6.016 6.008 6.021 6.023 6.009 6.010 6.018 6.015 6.013 Let us use 𝑥𝑖 to represent the readings, so 𝑥1 = 6.012mm, 𝑥2 = 6.016mm … 𝑥10 = 6.013mm. Studies show that when there are a large number of readings, however random the variation may seem, we can expect that roughly half of the readings are greater than the “true” value, and the other half are smaller; therefore, we can use the mean �̅� of all readings to represent the “true” value. In this case. �̅� = 𝑥1 + 𝑥2 + 𝑥3 … + 𝑥𝑁 𝑁 = 6.015mm In Microsoft Excel, you can call the function =AVERAGE() to facilitate the calculation of the mean �̅�. Each reading 𝑥𝑖 deviates from the mean �̅� by 𝑥𝑖 − �̅�. We define the standard deviation 𝝈 (Greek letter sigma) to represent how much variation exists. The formula to calculate 𝜎 is given below 𝜎 = √ (𝑥1 − �̅�) 2 + (𝑥2 − �̅�) 2 … + (𝑥𝑁 Page 1 3 2-Dimensional Forces Experiment objectives: 1. Use standard deviation to assess the quality of measurement data 2. Use average and uncertainty to assess reliability of measurement result 3. Distinguish experimental uncertainty from human error 4. Cultivate the habit of keeping all experimental data Experiment introduction: Force vector Vectors are quantities with directions. Force and velocity are good examples of vectors. When dealing with a vector in physics, one needs to take care of the direction. For example, as shown in Figure 1, there are two forces, �⃗�1 and �⃗�2, acting on a block. �⃗�2 is 10N pointing in the −𝑥 direction; �⃗�1 is 15N pointing at 𝜃 = 40° above the +𝑥 direction. If one needs to determine the net force, one has to break down the force vectors into their 𝑥 and 𝑦 components first. In this case, �⃗�2 = −10N�̂�. For �⃗�1, the following calculation can be used: �⃗�1 = 𝐹1,𝑥 �̂� + 𝐹1,𝑦 �̂� ⇓ 𝐹1,𝑥 = 𝐹1 cos(𝜃) 𝐹1,𝑦 = 𝐹1 sin(𝜃) } ⇒ { 𝐹1,𝑥 = 15 cos(40°) = 11.5N 𝐹1,𝑦 = 15 sin(40°) = 9.64N ⇓ �⃗�1 = 11.5N �̂� + 9.64N �̂� (1) Please note that the calculation follows the convention that a vector has an arrow on top of the symbol, and the magnitude of a vector has the symbol without the arrow. For example, �⃗�1 is a force vector; 𝐹1 is the magnitude of it. You are expected to follow the same convention. Figure 1: Forces on a block. To facilitate addition/subtraction of forces, the forces which do not line up with either 𝑥 or 𝑦 , in this case �⃗�1, must be broken down to their components. x y �⃗�1 �⃗�2 𝐹1,𝑥 𝐹1,𝑦 𝜃 Page 2 With the two forces in component form, �⃗�1 = 11.5N 𝑥 + 9.64N �̂� and �⃗�2 = −10N 𝑥, the net force on the box can be determined with vector addition, �⃗�total = �⃗�1 + �⃗�2 = (11.5N �̂� + 9.64N �̂�) + (−10N 𝑥) = 1.5N 𝑥 + 9.64N �̂�. Equilibrium In the study of motion and mechanics, equilibrium refers to the state of motion that the net force on an object is zero. Please note that “being in equilibrium” does not necessarily mean “being stationary”; an object moving with a constant velocity is in equilibrium because it undergoes no acceleration, which in turn means there is no net force on it. Please also note that the phrase “the net force on an object is zero” may refer to the following two different situations: 1. There is simply no force at all acting on the object. 2. There are multiple forces acting on the object, but they balance each other out. This situation happens very often. Experiment Procedures: In this experiment, you are expected to perform the following steps in order to determine an unknown force. 1. Use hanging masses to set up 3 known forces on a force table, 𝐹1 = 1.2~1.5N at 34°, 𝐹2 = 0.7~1.1N at 144° and 𝐹3 = 1.5~1.9N at 203°. 2. Use trial-and-error method to experimentally determine the magnitude and direction of the 4th force 𝐹4 which balances the existing 3 forces. 3. On a piecPhysics 1 Lab PHY 2221/2421 Section ____ Lab Title:__________________ Instructor:_________________ Name:____________________ Date:_____________________ Partners:__________________ 2D Forces– Data Analysis Data Analysis The same force table experiment is conducted to determine an unknown mass. The experiment setup is very similar to what you did in the classroom. The unknown mass is suspended on a string at the angle 90.0° (see picture), while two known masses are suspended on the other two strings at certain angles to balance the unknown one. The setup is shown in Figure 1. (a) (b) Figure 1: Force table experiment setup to determine an unknown mass. The side view (a) shows that one unknown mass and two known ones are suspended on strings to balance each other. In the top view (b), we can clearly read the directions of all forces. The unknown mass remains fixed, while the two known masses and their angles are adjusted to balance the unknown one. 10 sets of data were obtained and summarized in Table 1. Please use the data to answer the questions. Table 1: Force table experiment data. Unknown mass angle (°) 90.0 Data Count 1 2 3 4 5 6 7 8 9 10 𝑭𝟏 magnitude (gram) 396 301 251 482 669 739 478 653 261 397 𝑭𝟏 angle (°) 292.7 324.1 299.9 303.0 330.1 319.4 288.5 288.5 288.5 299.5 𝑭𝟐 magnitude (gram) 356 566 481 385 676 596 280 220 446 391 𝑭𝟐 angle (°) 244.5 244.5 255.1 226.6 211.5 200.0 235.6 197.5 260.0 240.5 Question 1: (5 points) Please only use data set 1, draw and label the 3 force vectors in the diagram below. Your drawing does not have to be to the scale, but should roughly reflect the directions of the 3 forces. Question 2: (5 points) Please only use the data set 1 to calculate the unknown mass. Show your calculation procedures and result below. Question 3: (5 points) Please use the 10 sets of data to calculate the unknown mass in Microsoft Excel. Record the result for each individual data below. Please note that in Excel the trigonometry functions take angles in radians. You need to convert angles given in degrees to radians first. The function =RADIANS() converts degree to radian; the function =DEGREES() does the opposite. Data count 1 2 3 4 5 6 7 8 9 10 Unknown mass (gram) Question 4: (5 points) Please process the 10 unknown mass results obtained in the previous question Average Standard Deviation Standard error Unknown mass (gram) x y 30 60 90 120 150 180 210 240 270 300 330 0 Question 5: (5 points) Please determine the 95\% confidence range for the “true” value of the unknown mass, show your calculation procedures; then report the final result in the format “mean value ± 95\% range”. Question 6: (5 points) The reference value of the unknown mass is 682 gram. Does the reference value fall in the 95\% confidence range of the mean value? What is the percent error? Lab 3 Report – Presentin