# Relation between way of measuring physical quantities

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

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.

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.