Absolute and Relative Error in measuring g: method 1

Name:

Lab Partners:

When entering numeric data, use exponentials: ie., 1.6 * 10^-19 = 1.6E-19.

Procedure

1. Place the steel ball in the release mechanism and measure the height "h" (in centimeters) from the bottom of the ball to the target pad. h should be between 20 and 60 cm. Be sure to measure the distance perpendicularly, and account for parallax by sighting the meter stick perpendiclarly to it and at the level of the ball.

   h₁ = cm

2. Turn the timer on and place the steel ball in the release mechanism. Push reset and release the ball. Record the time "t" (in seconds) for each of 10 trials.

   t1,1 = s				t1,6 = s
   t1,2 = s				t1,7 = s
   t1,3 = s				t1,8 = s
   t1,4 = s				t1,9 = s
   t1,5 = s				t1,10 = s

3. Repeat steps 1 and 2 for 3 different heights, for a total of 4 sets of 10 times.

   h₂ = cm

   t2,1 = s				t2,6 = s
   t2,2 = s				t2,7 = s
   t2,3 = s				t2,8 = s
   t2,4 = s				t2,9 = s
   t2,5 = s				t2,10 = s

   h₃ = cm

   t3,1 = s				t3,6 = s
   t3,2 = s				t3,7 = s
   t3,3 = s				t3,8 = s
   t3,4 = s				t3,9 = s
   t3,5 = s				t3,10 = s

   h₄ = cm

   t4,1 = s				t4,6 = s
   t4,2 = s				t4,7 = s
   t4,3 = s				t4,8 = s
   t4,4 = s				t4,9 = s
   t4,5 = s				t4,10 = s

   Note that if h is very small, the precision will suffer. Why?

Analysis

Do not round any values until you are told to do so!

1. For each of the 4 sets of data, calculate the average t:

   t_i,avg = ( t_i,1 + t_i,2 + t_i,3 + t_i,4 + t_i,5 + t_i,6 + t_i,7 + t_i,8 + t_i,9 + t_i,10 ) / 10

   t1,avg = s				t3,avg = s
   t2,avg = s				t4,avg = s

   together with the absolute and relative error in each set. To compute the absolute error, first find (for a given height) the difference between the largest time and the average, and the difference between the average time and the smallest. The absolute error is the larger of the two differences:

   t_i,abs = Maximum of ( t_i,largest - t_i,avg ) and ( t_i,avg - t_i,smallest )

   t1,abs = s				t3,abs = s
   t2,abs = s				t4,abs = s

   The relative error is then the quotient of the absolute error divided by the average, times 100 %:

   t_i,rel = 100 * t_i,abs / t_i,avg

   t1,rel = %				t3,rel = %
   t2,rel = %				t4,rel = %

   The errors will probably be different for each set of data.

2. For each of the 4 sets of data, calculate x = one half the square of the average t:

   x_i = t_i,avg² / 2

   x1 = s2				x3 = s2
   x2 = s2				x4 = s2

   For this experiment, the relative error in x equals the relative error in t. Calculate the absolute error in x by multiplying the relative error by x / 100 for each value of x:

   x_i,abs = t_i,rel * x_i / 100

   x1,abs = s2				x3,abs = s2
   x2,abs = s2				x4,abs = s2

   NOW round the values of x as follows, and use these rounded values in the subsequent calculations:

   if .01 < = x_i,abs < .1, round x_i to the nearest tenth
   if .001 < = x_i,abs < .01, round x_i to the nearest hundredth
   if .0001 < = x_i,abs < .001, round x_i to the nearest thousandth
   
   

   x1,rounded = s2				x3,rounded = s2
   x2,rounded = s2				x4,rounded = s2

3. The method of least squares can be used to find the equation of the line which best fits the experimental data. This technique is called linear regression. If x_i and y_i are the measured x and y values, there are n observations (so i goes from 1 to n), and X and Y are the averages of the x_i and y_i, respectively, then the slope m and x-intercept b of the best linear fit are given by:

   m = (n X Y - S x _i y _i ) / (n X ² - S x _i ²)

   b = Y - m X

   where the S denotes a sum. For instance, with the following data for x and y:

   (0.5, 5) (0.4, 4) (0.3, 3) (0.2, 2),

   n = 4, X = 0.35, Y = 3.5, S x _i y _i = (2.5 + 1.6 + 0.9 + 0.4) and S x _i ² = (0.25 + 0.16 + 0.09 + 0.04). These values give

   m = ((4 * 0.35 * 3.5) - 5.4) / ((4 * 0.1225) - 0.54)

   = 10 (surprise!)

   and

   b = 3.5 - 10 * .35 = 0.

   Calculate m and b for y = h and x = x_rounded, to obtain the slope and intercept (Do not round anything in this calculation until the end; the denominator will be very small and so the final result is very sensitive to premature rounding):

   X = ( x₁ + x₂ + x₃ + x₄ ) / 4

   X = s²

   Y = ( h₁ + h₂ + h₃ + h₄ ) / 4

   Y = cm

   m = ( 4 * X * Y - ( x₁ * h₁ ) - ( x₂ * h₂ ) - ( x₃ * h₃ ) - ( x₄ * h₄ ) ) / ( 4 * X ² - x₁² - x₂² - x₃² - x₄² )

   m = cm / s²

   What do you think is the "correct" value for m and why?

   b = Y - m * X

   b = cm

   What do you think is the "correct" value for b and why?

4. Plot your data points ( x, h ) and the least squares fit line (using "m" and "b" from the last step) on the same graph. By hand, try to draw a little box of height twice the absolute error in h (probably 0.1 cm) and width twice the absolute error in x, centered on each data point:
   

   Depending on the scale of your vertical axis, the boxes may be too short to draw; in this case, draw a horizontal error bar for each point, from x_i - x_{i, abs} to x_i + x_{i, abs}.

   If the fit line intersects all of the boxes, your data is consistent with the fit line. If the line does not fit well, the situation is more obscure. Note that large error boxes correspond to environmental or procedural errors, while small error boxes which are inconsistent with the results of the linear regression indicate "built-in" error in the apparatus. What do you conclude from your data?