I am stuck with the following question. I am not sure how to proceed with the following question.

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I am not sure even how to proceed with the first step. What I know is the concept of relative percentage error, which is $\dfrac{\Delta x}{x} \times 100%$. So, let's say even if I set the time period to be $T$ and student A measures it like $T_1, T_2,... T_{N_A}$, then time period he measure will be $\left \langle T \right \rangle =\dfrac{T_1+T_2+..+T_{N_A}}{N_A}$.

I think uncertainty for student A would be $$\sqrt{\dfrac{\displaystyle{\sum_{i=1}^{N_A} (\left \langle T \right \rangle -T_i)^2}}{N_A} }$$

Now, I am unsure how to make a similar argument for student B's measurement. More importantly, I am getting no clue how to approach further in this problem.

Let B measures $T_{N_B}$ time for $N_B$ oscillations, so he would measure $\dfrac{T_{N_B}}{N_B}$ time as "time-period".

Now, I am not sure how to determine uncertainty in the case of student B.

Idea is to find uncertainty and make them equal to find the relation between $N_A$ and $N_B$ but how to get it? Any help would be appreciable. Thanks.


1 Answer 1


You can think about it this way:

There is a time measurement, $t$ [s], with some uncertainty, say $\delta t=\pm 0.1$ s, which is fixed. So if the student measures the duration of 1 oscillation, the uncertainty is $\pm 0.1$ s. If the student measures the cumulative duration of 1000 consecutive oscillations, the uncertainty is still $\pm 0.1$ s. But in the latter case, the total measured time is 1000 times longer than in the former case, so the uncertainty as a proportion of the total time is 1000 times lower.

Now, the period for a single measurement is calculated as $T=t/N$, where $N$ is the number of periods. Let's assume that the error in $N$ is zero. So the uncertainty in $T$ for a single measurement, $\delta T_1$, is going to be $$ \delta T_1=\delta t\frac{\partial T}{\partial t}=\frac{\delta t}{N}. $$ Clearly, the uncertainty decreases linearly with the number of consecutive oscillations measured in the manner of student B. This is because the (fixed) error becomes a linearly diminishing proportion of the total measured time when more oscillations are measured. Increasing $N$ is directly increasing the signal-to-noise ratio (SNR), which proportionally improves uncertainty.

Student A's measurement, on the other hand, only ever uses a single oscillation at a time. So student A's single-measurement uncertainty will be $\delta T_1 = \delta t$. To improve uncertainty from here, student A repeats the measurement $N_A$ times, which leads to a total uncertainty of $$ \delta T_{N_A}=\frac{\delta T_1}{\sqrt{N_A}}=\frac{\delta t}{\sqrt{N_A}}, $$ as you know. This relation is different from student B's uncertainty because student A is relying on statistical averaging to improve the uncertainty rather than directly increasing the SNR.

The moral of the story is always try to improve SNR first. Begin to average only when the practical limit of SNR is reached.


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