Suppose a student recorded the following data during Step 10 of this experiment, A 6.2 deg 4.7 deg 3.2 deg 2.4 deg 1.8 deg
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Suppose a student recorded the following data during Step 10 of this experiment,

A 6.2 deg 4.7 deg 3.2 deg 2.4 deg 1.8 deg
Time 2 s 4 s 6 s 8 s 10 s
The student was careful to start taking data when the angular position of the pendulum was zero. Unfortunately he does not have a computer available to fit the data as you will in Step 12, but he knows that the amplitude of oscillations decays exponentially according to A= A0e-(Bt)+ C, where C = 1. Use the data obtained by the student to plot a straight line graph of the appropriate variables on the graph paper, and use it to find values for the parameters A0 and B.

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  1. Answer:

      B = -0.215 s⁻¹,  A₀ = 1.537º

    Explanation:

    To be able to make this graph we must linearize the data, the best procedure is to calculate the logarithm of the values

               A = A₀ [tex]e^{-Bt}[/tex] + C

    The constant creates a uniform displacement of the graph if we start the graph at the angle of 1, the constant disappears, this is done by subtracting 1 from each angle, so the equation is

                (A-1) = A₀  e^{-Bt}

    We do the logarithm

               Log (A-1) = log Ao –Bt log e

    Make this graph because paper commercially comes in logarithm 10, if we use graph paper if we can calculate directly in base logarithm e, let’s perform this calculation

                  Ln (A-1) = Ln A₀ – Bt

    To graph the points we subtract 1 from each angle and calculate the logarithm, the data to be plotted are

                  θ’= ln (θ -1)

    θ(º)    t(s)        θ'(º)

    6.2      2     1.6487

    4.7      4     1.3083

    3.2      6    0.7885

    2.4      8    0.3365

    1.8     10   -0.2231

    We can use a graph paper and graph on the axis and the primary angle (θ’) and on the x-axis time, mark the points and this graph is a straight line, we see that the point for greater time has a linearity deviation , so we will use the first three for the calculations

    To find the line described by the equation

                  y -y₀ = m (x -x₀)

                 m = (y₂ -y₁) / (x₂ -x₁)

    Where m is the slope of the graph e (x₀, y₀) is any point, let’s start as the first point of the series

                (x₀, y₀) = (2, 1.65)

               (x₂, y₂) = (2, 1.65)

               (x₁, y₁) = (6, 0.789)

    We use the slope equation

               m = (1.65 – 0.789) / (2-6)

               m = -0.215  

    The equation is

               y – 1.65 = -0.215 (x- 2)

               y = -0.215 x +0.4301

    We buy the two equations and see that the slope is the constant B

                B = -0.215 s⁻¹

    The independent term is

                b = ln A₀

                A₀ = [tex]e^{b}[/tex]

               A₀ = [tex]e^{0.43}[/tex]

               A₀ = 1.537º

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