Time — Question 4

Question 4

Twenty-one crossings do not make twenty-one periods

An oscillator passes a fixed mark in the same direction once per period. Its period $P$ is constant for this model. An observer starts timing at one such crossing and stops at the twenty-first crossing in the list, counting the start crossing as number 1. The displayed total is $M=30.00\ \mathrm{s}$.

Calibration establishes that the elapsed interval between the chosen first and last crossings is in $[M-0.20,M+0.20]\ \mathrm{s}$. For this exercise, assume this total timing bound remains $\pm0.20\ \mathrm{s}$ even if more periods are timed; other errors are negligible.

1. How many complete periods were timed? Find the central estimate and hard interval for $P$.

2. A student divides $30.00\ \mathrm{s}$ by 21. Explain the counting error using the simplest possible two-crossing example.

3. How many complete periods must be timed to obtain a period half-width no greater than $0.0030\ \mathrm{s}$ under the stipulated error model? How many crossings, including the first, must be counted?

4. Using the central period estimate, find the approximate duration of that longer observation. Explain two reasons why increasing the number of periods might eventually stop improving an actual experiment as predicted by this ideal model.

Question 4 - Solution

Twenty-one crossings do not make twenty-one periods

(a) Count the gaps between the markers.

Twenty-one successive same-direction crossings enclose $21-1=20$ complete periods. Therefore $$P_0=\frac{30.00\ \mathrm{s}}{20}=\boxed{1.500\ \mathrm{s}},\qquad P\in\left[\frac{29.80}{20},\frac{30.20}{20}\right]\ \mathrm{s} =\boxed{[1.490,1.510]\ \mathrm{s}}.$$ The half-width is $0.010\ \mathrm{s}$.

(b) Markers bound intervals; they are not intervals.

Two consecutive same-direction crossings mark the beginning and end of one period, not two. In general, $k$ listed crossings enclose $k-1$ periods. Dividing by 21 would give approximately $1.42857\ \mathrm{s}$, an underestimate caused by an extra assumed period.

(c) Divide the total bound by the number of periods.

Timing $n$ periods gives a period half-width $0.20\ \mathrm{s}/n$. The requirement is $$\frac{0.20}{n}\le0.0030,\qquad n\ge66.\overline{6}.$$ Thus the smallest integer choice is $\boxed{67\text{ periods}}$, requiring $\boxed{68\text{ crossings}}$ including the start. Its half-width is approximately $0.002985\ \mathrm{s}$; 66 periods would give about $0.003030\ \mathrm{s}$ and fail the requirement.

(d) Check the assumptions behind a longer observation.

$$\Delta t_{67}=67P_0=100.5\ \mathrm s,\qquad \delta P_{67}=\frac{0.20}{67}\ \mathrm s\approx0.002985\ \mathrm s.$$

The central predicted duration is $67(1.500\ \mathrm{s})=\boxed{100.5\ \mathrm{s}}$.

A real oscillator’s period could drift during the longer run, making the quotient a run-average rather than one fixed period. A timer’s rate error may also accumulate with duration instead of contributing a fixed total bound. Missed crossings or changed trigger conditions can add further errors. Longer timing helps as stated only while the fixed-period, correct-count, and fixed-total-bound assumptions remain valid.