For questions that require calculations, please show all formulas used and detailed algebra steps concluding with the nal numerical result. For questions that require text, please use complete sentences. This prelab is just like the previous prelab, but with a different set of numbers. This prelab and the one prior to this prelab will be helpful practice for your Pendulum Lab reports. Before doing this prelab, go to the "Files" link on Canvas and read the document "Logarithmic Plots". Use this document to help you work on this prelab exercise. A group of scientists are running a series of experiments and collecting data from two measurable quantities, A and B. They suspect that these quantities are related by a power law equation of the form A = cB\Introduction We should all remember the generic equation for a straight line: y = mx + b (Eq. 1) which says that the dependent variable y is proportional to the independent variable xwith a constant of proportionality m (which is the slope of the line) and the y-intercept is b. If the relationship between two sets of data is linear then Equation 1 describes how those data relate to each other. Figure 1 below shows an example of two sets of data (x and y values) that are linearly correlated to each other. 6 Dependent variable, v U o 1 2 3 4 5 s 7 Independent variable, it Figure 1. A set of data correlated by Equation 1 where the value of the slope is equal to 1 and the y intercept is equal to -1. Any time a graph is a straight line, then this generic form y = mx + b describes the relationship between the dependent variable on the vertical axis and the independent variable on the horizontal axis. Note: having the dependent variables on the vertical axis and independent variables on the horizontal axis is a mathematical convention, not a physical one. One set of data may physically depend on another, but as you will see in this lab these data can be manipulated and graphed many different ways. So the mathematical convention of a dependent variable may not match the \"physical dependent variable\" of the experiment. Logarithmic plots Earlier this quarter a set of experiments were done predicated on the following equation: by = gatz. If 'Ay' vs. 'tz' was plotted you could determine the acceleration of the object in freefall. However, what if you didn't know the exponent on the time variable? Would you still be able to determine the acceleration? There is a graphical method that can be used to determine the exponent on the independent variable and the coefcient in front of the independent variable. This method will be used in the upcoming lab report. To illustrate how logarithmic plots work and how we can use them to find exponents and coefficients, let's use the previous example of objects in free fall. Ay = =at2 (Eq. 2) Now, if we take the natural log of both sides of this equation, we get (In Ay) = 2(Int) + In (2) (Eq. 3) This means we can collect 'Ay' and corresponding 't' values, take the natural log of these values and plot: In(Ay) vs. In(t) and the slope would be 2 (the exponent on the independent variable of time) and the intercept would be the natural log of 'a/2' (the natural log of the coefficient on the independent variable). Let's say a student took the following data for an object at various drop heights Drop height (m) Drop time (s) Drop 1 0.2 0.36 Drop 2 0.5 0.58 Drop 3 0.8 0.73 Drop 4 1.1 0.85 Drop 5 1.4 0.97 Drop 6 1.7 1.06 Drop 7 2.0 1.16 The student then takes the natural log of both sets of data to get: In [Drop height] Ln [Drop time] (In m) (In s) Drop 1 -1.609... -1.021... Drop 2 -0.693... -0.544... Drop 3 -0.223... -0.314... Drop 4 0.095... 0. 162... Drop 5 0.336... -0.030... Drop 6 0.530... 0.058... Drop 7 0.693... 0. 148... Note: the "..." indicates that the fractional value has more digits than are listed in this table.Then the plot of I In(Ay) vs. In(t) would look like this: Ln(Ay) vs. Ln(t) Ln(Ay) -1.2 -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4 -0.5 -1 -1.5 Ln(t) If the student used MS Excel or curve.fit to find the slope and intercept of this plot, the student would get: slope = 1.97... and an intercept of 0.403.... Squares and square root relationships between variables in physics are fairly common. Thus, it would be a safe assumption that the exponent on the independent variable (in this example, time) is squared (not raised to the 1.97 power) because there is some error in the data. What about the intercept? Based on the information provided in Eq. 3, the intercept would be the natural log of the acceleration of the object divided by 2. Knowing this we can extract information about the acceleration by doing the following First step, setup Intercept = In(a/2) Second step, take the exponential of both sides e (intercept) = a/2 Third step, multiply by 2 a = 2* e(intercept) This would yield a value of 2.9926... most likely the acceleration is 3.0 m/s2 when error is taken into consideration