Time — Question 9

Question 9

Which clock is stable, and which ends on time?

Two clocks are synchronized to a reference at $t=0$. During each successive reference interval of exactly $100.000\ \mathrm{s}$, they advance by the amounts shown. Assume each clock’s rate is constant within each listed interval, and take all numbers as exact for this comparison. For the successive reference intervals $0$–$100$, $100$–$200$, $200$–$300$, and $300$–$400\ \mathrm{s}$, clock A advances $100.010\ \mathrm{s}$ each time. Clock B advances $100.040$, $99.960$, $100.040$, and $99.960\ \mathrm{s}$, respectively. Define the offset $E(t)$ as clock reading minus reference time. For this question, compare rate repeatability by whether equal reference intervals produce the same clock advance; no statistical stability measure is required.

1. Find $E(t)$ for each clock at $t=100,200,300,400\ \mathrm{s}$. Which has more repeatable interval advances, and which agrees with the reference at the final endpoint?

2. Find the largest absolute offset attained by each clock during the full 400 s, using the stated constant-rate behavior within intervals. Does final agreement imply the best timing throughout the record?

3. Suppose each full clock reading is multiplied by one constant chosen to make the corrected clock agree at both $t=0$ and $t=400\ \mathrm{s}$. Determine that factor for each clock and the resulting maximum absolute offset over this record.

4. What would be needed before claiming the observed behavior or correction will continue in a future run?

Question 9 - Solution

Which clock is stable, and which ends on time?

(a) Accumulate the interval gains and losses.

$$E(t_j)=\sum_{i=1}^j(\Delta C_i-100.000\ \mathrm s).$$

$$\begin{aligned}(E_A(100),E_A(200),E_A(300),E_A(400))&=(.010,.020,.030,.040)\ \mathrm s,\\(E_B(100),E_B(200),E_B(300),E_B(400))&=(.040,0,.040,0)\ \mathrm s.\end{aligned}$$ Clock A has the more repeatable rate over these intervals: every advance is identical. Clock B agrees with the reference at 400 s, while A is 0.040 s fast. Repeatability and endpoint accuracy answer different questions.

(b) Check the whole interval, not only the last timestamp.

Within each interval the offset changes linearly, so its extreme values occur at interval endpoints. Both clocks reach maximum absolute offset $\boxed{0.040\ \mathrm{s}}$. B’s zero final offset conceals its earlier excursions. Final agreement is therefore not evidence of uniformly small timing error.

(c) Correct a consistent rate, then test the residuals.

For A the required factor is $$k_A=\frac{400.000}{400.040}=\boxed{\frac{10000}{10001}}.$$ Because A’s rate is constant throughout this model, $C_A(t)=1.0001t$. The factor removes its rate error at every time, giving zero residual offset over the record.

B already reads 400.000 s at the final endpoint, so its required factor is $\boxed{k_B=1}$. The alternating rates remain unchanged, and its maximum absolute residual offset stays $0.040\ \mathrm{s}$. One overall rate correction cannot remove that within-run variation.

(d) A finite calibration record is not a guarantee of the future.

Repeat comparisons over the relevant durations and operating conditions, and establish bounds on changes in rate or offset. The exact conclusions above apply to the supplied piecewise-constant model and record. They do not, without further evidence, certify future stability, temperature response, or long-term accuracy.