Absolute and Relative Error in measuring g: method 2

The velocity of a falling object is

v = g t.

We will "measure" g in another way, and compare the precision of our measurement with that from the previous experiment.

The apparatus for this experiment uses a spark timer, which develops voltages in excess of 30,000 volts. The instructor will demonstrate its usage and closely supervise you as you perform the experiment. At no time during the experiment should you approach the freefall apparatus while the timer is turned on.

The spark timer generates a spark every 1/60th of a second. The weight, which has a conducting ring around it, falls between a wire and a spark tape, which is backed by a grounded pole. The spark tape is a special type of paper which will clearly register the spark between the wire and the pole. As the weight falls, every 1/60th of a second the spark burns a tiny hole in the spark tape, providing a record of the distance travelled in each interval.

Name:

Lab Partners:

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

Procedure

Perform the experiment as directed by the instructor. It is convenient to number the sparks, starting with zero. Expect 30 sparks, although the instructor may choose a subset of your sample if some are faint or missing. Using a meter stick, measure the vertical distance di,i+1 between each pair of sparks:

d0,1 = cm d15,16 = cm
d1,2 = cm d16,17 = cm
d2,3 = cm d17,18 = cm
d3,4 = cm d18,19 = cm
d4,5 = cm d19,20 = cm
d5,6 = cm d20,21 = cm
d6,7 = cm d21,22 = cm
d7,8 = cm d22,23 = cm
d8,9 = cm d23,24 = cm
d9,10 = cm d24,25 = cm
d10,11 = cm d25,26 = cm
d11,12 = cm d26,27 = cm
d12,13 = cm d27,28 = cm
d13,14 = cm d28,29 = cm
d14,15 = cm d29,30 = cm

Analysis

  1. Compute the total distance fallen at each point:

    hj = the sum of all di,i+1 from i = 0 to i = j-1

    (ie., h3 = d0,1 + d1,2 + d2,3)

    h1 = cm h16 = cm
    h2 = cm h17 = cm
    h3 = cm h18 = cm
    h4 = cm h19 = cm
    h5 = cm h20 = cm
    h6 = cm h21 = cm
    h7 = cm h22 = cm
    h8 = cm h23 = cm
    h9 = cm h24 = cm
    h10 = cm h25 = cm
    h11 = cm h26 = cm
    h12 = cm h27 = cm
    h13 = cm h28 = cm
    h14 = cm h29 = cm
    h15 = cm h30 = cm
    Plot the total distance fallen at each point versus time (the time coordinate for hj should be t = j / 60 sec). Recalling the equation for h in the previous experiment, does this seem to be a reasonable plot of h? Why?

  2. Divide the distance between each pair of points by the time interval between those points to get the average velocity between those points:

    vi,i+1 = di,i+1 / (1 / 60)

    v0,1 = cm / s v15,16 = cm / s
    v1,2 = cm / s v16,17 = cm / s
    v2,3 = cm / s v17,18 = cm / s
    v3,4 = cm / s v18,19 = cm / s
    v4,5 = cm / s v19,20 = cm / s
    v5,6 = cm / s v20,21 = cm / s
    v6,7 = cm / s v21,22 = cm / s
    v7,8 = cm / s v22,23 = cm / s
    v8,9 = cm / s v23,24 = cm / s
    v9,10 = cm / s v24,25 = cm / s
    v10,11 = cm / s v25,26 = cm / s
    v11,12 = cm / s v26,27 = cm / s
    v12,13 = cm / s v27,28 = cm / s
    v13,14 = cm / s v28,29 = cm / s
    v14,15 = cm / s v29,30 = cm / s

    Assume that the average velocity between any two data points (vi,i+1) occurred at the halfway point between them; this "average time" is:

    ti,i+1 = i / 60 + 1 / 120 s.

    In the figure below, the odd multiples of 1/120 s are the average times:

    t0,1 = s t15,16 = s
    t1,2 = s t16,17 = s
    t2,3 = s t17,18 = s
    t3,4 = s t18,19 = s
    t4,5 = s t19,20 = s
    t5,6 = s t20,21 = s
    t6,7 = s t21,22 = s
    t7,8 = s t22,23 = s
    t8,9 = s t23,24 = s
    t9,10 = s t24,25 = s
    t10,11 = s t25,26 = s
    t11,12 = s t26,27 = s
    t12,13 = s t27,28 = s
    t13,14 = s t28,29 = s
    t14,15 = s t29,30 = s

    Using the method of linear regression with y = average velocity and x = average time, calculate the slope and intercept of the velocity vs. time graph; n is the number of sparks:

    X = ( t0,1 + ... + t29,30 ) / n

    X = s

    Y = ( v0,1 + ... + v29,30 ) / n

    Y = cm / s

    m = ( n * X * Y - ( t0,1 * v0,1 ) - ... - ( t29,30 * v29,30 ) ) / ( n * X 2 - t0,12 - ... - t29,302 )

    m = cm / s2

    b = Y - m * X

    b = cm / s

    Plot the fit line on a graph of your (t, v) data. What can you conclude?

  3. Compare your results with those from the last experiment. Is one method of measuring g more accurate than the other? Why?

©2004, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.

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