Answer the following questions about precision and accuracy for measuring the period of a pendulum in lab?
#a)# If a stopwatch only reports time in seconds to one decimal place, to how many decimal places can the period of a pendulum be reported? What assumptions are made when recording 8 periods in a row without holding the pendulum in between?
#b)# How can the accuracy of the #8# periods in #(a)# be improved? What should be reported for the period after measuring it those #8# times, and how should it be reported?
#c)# If another student found that #100# swings took #"92.8 s"# , what might the average period be? Is this a good approximation? Why or why not? If not, how can these numbers be improved?
1 Answer
No. The stopwatch has only one decimal place, and thus the period can only be stated to the same degree of uncertainty, i.e. to one decimal place.
Furthermore, no description of any nonidealities of each of the
#8# pendulum periods were given (such as someone bumping it, or the friction of the pendulum's string on what it's hanging off of), so we would be assuming:
- that it was
#8# uninterrupted swings.- that each swing was identical with NO friction (rather than of slightly decreasing periods each time due to friction).
It would be more accurate to do
#8# trials of letting the pendulum swing for one period and timing that ONE period EIGHT times.Then, the average should be taken of each of the
#8# periods recorded, and the standard deviation calculated. A larger sample of multiple trials is always better than doing one trial.Finally, the average period along with its uncertainty recorded. An example is if you have these
#8# trials:
#"Trial"" "" ""Recorded Period (s)"#
#" "1" "" "" "" "" "0.9#
#" "2" "" "" "" "" "0.8#
#" "3" "" "" "" "" "1.1#
#" "4" "" "" "" "" "0.9#
#" "5" "" "" "" "" "0.8#
#" "6" "" "" "" "" "0.8#
#" "7" "" "" "" "" "1.0#
#" "8" "" "" "" "" "0.9# The average would then be
#(0.9 + 0.8 + 1.1 + 0.9 + 0.8 + 0.8 + 1.0 + 0.9)/8 = "0.9 s"# Its sample standard deviation would be:
#s = sqrt((sum_(i=1)^(8) (x_i - barx)^2)/(N-1))#
#= sqrt(((0.9 - 0.9)^2 + (0.8 - 0.9)^2 + (1.1 - 0.9)^2 + (0.9 - 0.9)^2 + (0.8 - 0.9)^2 + (0.8 - 0.9)^2 + (1.0 - 0.9)^2 + (0.9 - 0.9)^2)/(8-1))#
#= 0.1_(069) " s" -> "0.1 s"# (one decimal place)And thus, the average period should be reported, as:
#color(blue)(0.9)# #color(blue)((pm 0.1))# #color(blue)("s")#
If another student found
#100# swings took#"92.8 s"# , then he or she can pretend that all the swings are identical (which is tacitly untrue!) and claim that the period is:
#"92.8 s"/("100 swings") ~~ "0.9 s/swing"# which is of course an estimate, because by the 50th swing you should imagine the pendulum to have significantly slowed down (thus rendering each period not identical to the previous and the first 50 swings to be more accurate than the latter 50).
No standard deviation can be recorded from such data, so to be practical, the student should redo the experiment completely, and do only 10 swings maximum, recording the period each time...
