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.
- 天同Lv 710 年前最愛解答
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.