Why is the value for he period T more precise if u measure 20 swings(20T) at once?

Answer 1

Because, given an absolute error E of, say 0.1 sec (typical of a stop watch), the percent error R in measuring n swings is (E/(nT))*100%. R drops in proportion to n.

Answer 2

OK, assume you measure t = T + e; where T is the true period and e is some error, which is small compared to T. In math talk e<<
Now you measure each period and do that N times. The sum of your measures is S = sum(t) = sum(T + e). The average tavg = S/N = sum(T + e)/N = sum(T)/N + sum(e/N) = T + eavg. In measuriing the period each time and then taking the average, we are subject to an eavg error in our conclusion. But the central limit theorum says that eavg = 0; so that we have tavg = T, the true T.

Now you swing the pendulum N times and measure the total time ttot it took for N swings. In which case ttot = (T + e1) + (T + e2) + ... + (T + eN). If we assume the errors are about the same in value, then ttot = N(T + e); so that t = ttot/N = T + e. Thus, the derived t from this test is simply the true period T plus some e <<< T. Thus, by measuring a total period and dividing through by the number of periods to get a derived measure, you are sitll left with an error e.

In the end, it's better to take multiple (N) measures, sum them up, and divide through by N to get the average. That results, because the inherent errors in each measure tend to cancel out.

Answer 3

Because when you measure 20 times period of one swing the errors of individual measurements are uncorrelated, and averaged value of 20 measurement is only √20 better than single measurement.

But if you measure all 20 swings at once, the error in determination of the beginning of one swing exactly cancels the error in determination of the end of previous swing (i.e. errors are 100% anti-correlated) and the accuracy improves 20 times.