Phy31 Lab

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  • Topic: Complex number, Exponentiation, Polynomial
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Lab 2

Physics 190

Acceleration “g” Due to Gravity – Method 2
Introduction Tonight we will measure the acceleration due to gravity again. This time however, we will collect more data and the analysis will be different. We will first fit the data using a second order polynomial. Recall for a mass falling from rest, that 1 (1.1) y  a yt 2 2 Suppose a mass falls through n successively greater displacements, each time starting from rest. The displacements can be expressed a 2 y  y t ;   1 n  . (1.2) 2 Analyzing the Data Data for y is not linear in time t. We have two unique ways we can analyze the data. The first is to simply plot the data with vertical displacement on the y-axis and time on the x-axis and perform a 2nd order polynomial curve fit. We can then extract acceleration from the coefficient of the 2nd order term. The second method involves transforming the nonlinear data into a linear form by means of the logarithm from which we can extract acceleration. We are going to use both methods because it demonstrates the power of mathematics as a data analysis tool. Fitting the Data to a 2nd Order Polynomial Free-fall data is shown in figure 1 and has the form y  At2  Bt  C

(1.3)

Figure 1. Free-fall plot (dots) and 2nd order fit (solid line).

If we fit ideal free-fall data to equation (1.3) we should find that B = 0, C = 0, and A = ay/2. If you look at the polynomial fit equation embedded in figure 1 you will see

BWhitecotton

Page 1 of 7

Lab 2

Physics 190

that B = -10-13, C = -10-14, and A = -4.905. So the data is not perfect but essentially both B and C are zero while A = -4.0905. If you compare the polynomial equation to our kinematic equation… y  At 2  Bt  C a y  y t 2  vyit  yi 2 …it becomes immediately evident that B corresponds to initial velocity, C the initial position, and A = ay/2. If dropped from rest, initial velocity and position are zero. This all boils down to the fact that fitting a second order polynomial to free-fall data should provide the acceleration due to gravity directly. Simply plot displacement (yaxis) vs. time (x-axis) and use Excel, Vernier, calculator, or any tool that will perform a polynomial fit of order 2. Then ay = 2A which in the example above gives ay = 2(-4.905) = -9.81. Using the Logarithm to Linearize Data and Fit We begin with equation (1.2), generalize and take absolute value ay m y  t . 2 Vertical in figure Time Equation (1.4) is plotted as data belowDisplacement vs2. 25

(1.4)

20

|y(t)| (m)

15

10

5

0 0 0.5 1 t (sec) 1.5 2 2.5

Figure 2. Absolute value of vertical displacement versus freefall time.

Taking the log we obtain
 ay  . log  yn   m log tn   log   2    mXn Y n

(1.5)

B

Equation (1.5) has the slope-intercept form of a line. Plotting the log of the data of figure 2, we obtain figure 3. The curve fits a straight line that has the form of Y = mX + B with m = 2.0108 and B = 0.6896.

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Lab 2

Physics 190

Linearized Data
1.5

y = 2.0108x + 0.6896 R2 = 1

1

0.5

Log( |y(t)| )

0 -1.2 -1 -0.8 -0.6 -0.4 -0.2 -0.5 0 0.2 0.4

-1

-1.5 Log(t)

Figure 2. Linearized data from figure 1 data above.

Recalling that B = log(|ay|/2) = 0.6896, we can solve for the acceleration ay. Inverting we get ay  100.6896 2 ay  4.893 . 2 a y  9.787 Recall that our lab is at latitude  = 32.745°. Therefore the acceleration due to gravity in our lab should have magnitude g  9.795 .

Computing experimental error we find

a

y

 g g

 100%  9.787  9.795 100%  0.0863% . 9.795

This is quite respectable but also uncharacteristically low for experiments in our lab. This experiment, if carefully done, can yield 1% error.

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Lab 2 Procedure

Physics 190

Set up the apparatus as we did last week. See figure 3 below for typical arrangement – this should look familiar. Spherical mass

to= 0 s...
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