Mass-spring oscillation

1. The graph does not pass through the origin owing tothe mass of the spring and the movement of its various parts. Derive anexpression to show that the intercept on the m-axis gives the effectivemass of the spring.

2. Suppose the smallest division on the scale of thetiming device is td, thereaction time of a normal person is tr,which one should be used for estimating the uncertainty on the measurement ofperiod T after a single timingactivity? Explain briefly.

1 個解答

  • 天同
    Lv 7
    10 年前

    Since you did not give a simple description to your experiment and what dependent and independent variables are plotted on the graph, I suppose you are plotting the equare of period of oscillation (T^2) against the mass hung on the spring (m)

    If the effective mass of the spring (Me) is taken into account, the equation describing the oscillation is given by,

    (m + Me)a = -k.x

    where k is the spring constant and x is the displacement of the oscillating mass

    hence, a = -[k/(m + Me)]x

    The square of the angular frequency of oscillation w^2 = [k/(m + Me)]

    The period square T^2 is then given by,

    T^2 = (4.pi)^2.[(m + Me)/k]

    i.e. T^2 = [(4.pi)^2/k].m + [(4.pi)^2/k].Me

    Hence, a plot of T^2 against m gives a straight line not passing through the origin.

    The intercept on the mass-axis (x-intercept) is found by setting T^2 = 0

    we have, 0 = [(4.pi)^2/k].m + [(4.pi)^2/k].Me

    i.e. (m + Me) = 0

    or Me = -mThus, the intercept on the mass-axis gives the effective spring mass

    2. This depends on how precise the timeing device is, i.e. how small is td compared to tr. If tr > td, then tr should be used for uncertainty estimation.

    2011-02-07 17:18:23 補充:

    As a supplement to Q(2), the uncertainty in T = tr + td, hence the one with larger value should dominate the uncertainity.