The lab we have just finished compared an object’s gravitational mass (obtained on our triple beam balances) with its inertial mass (obtained by examining its period on an inertial balance). The actual formula for this relationship is where **T** is the period measured in seconds, **m** is the mass vibrated on the spring measured in kilograms, and **k** is the spring constant (a measure of how must force is to displace a spring from its equilibrium position; that is, the spring's stiffness) measured in newtons/meter. This formula is actually only true for a massless spring; hence the need to subtract away the behavior of the empty pan (which obviously had mass) during our analysis. Had we graphed **Period vs. Mass **for our class data, the graph would have been a power function which, although it is a function, is not easily manipulated. A power function just means that the independent variable is raised to a power. Also recall from algebra that a graph is always titled as **y vs x** – that is, as the y-axis variable vs the x-axis variable. The exponent on our graph is **0.5448**. If our data had been perfect, this exponent should have been 0.5 or one-half. Our correlation coefficient, R^{2}, is 0.9985 and represents how well our data conforms to our trend line; that is, how well the data “fits together.” An R^{2} = 1 is perfect. To linearize our data, we plotted **T**^{2} vs M. Linearizing makes the relationships in the data easier to comprehend since a line represents a direct proportion between the variables T^{2} and M instead of the previous power function between T and M. A graph of Period Squared (for mass alone) vs Vibrated Mass Notice that the correlation coefficient is not quite as good, it has been reduced to 0.9973 from 0.9985. This is to be expected when the data has been through several additional rounds of calculations. The safest method of doing multiple calculations with numerical data is to let the running decimals stay on your calculator screen and only round off the FINAL answer. Repeated rounding of intermediate answers introduces errors. If we wanted to determine the stiffness, or elasticity, of the two steel bands of the inertial balance (**k** in the original formula), we would do this by setting the numerical value of the slope from our line of best fit (0.7835 sec^{2}/m) equal to the collection of variables composing the coefficient of **m** (the expression in parentheses) and solve for **k**. Remember that mass is being graphed on the x-axis since it was the independent variable during the experiment. Don’t worry about the units on **k**, we will cover them later. Right now all I want you to recognize is how we were able to set the expression representing the slope from the formula (4π^{2}/k) – the coefficient of the x-axis variable – equal to the numerical value for the slope (0.7835) to solve for the only unknown in the equation, the stiffness of the spring, **k**. This technique is called *linear regression theory *and it will become a common tool that is used during labs. |