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## Homework Statement

Measuring the period of a sine wave using oscilloscope, for the best accuracy, is it better to measure only one cycle?

## Homework Equations

## The Attempt at a Solution

Let [itex]t=T_{1}+T_{2}+\cdots + T_{n}[/itex] (Measuring n cycle)

From error propagation formula,

[itex]\delta t = \sqrt{(T_{1})^2+(T_{2})^2+\cdots + (T_{n})^2}[/itex] (or [itex]\delta t= n \delta T_{1} [/itex] ???)

but as [itex] T_{1}, T_{2}, \cdots , T_{n}[/itex] are independent so the value of each [itex]\delta T_{n} [/itex] is same. So, let [itex] \delta T= \delta T_{1} = \delta T_{2}= \cdots = \delta T_{n} [/itex]

This [itex] \delta T [/itex] is a error for a period.

Therefore,

[itex] \delta t= \sqrt{n(\delta T)^{2}}=\sqrt{n} \delta T[/itex]

But, [itex] \delta t \propto n[/itex] (as n increases, the scale for one div the screen become smaller,

which increase your error)

So [itex] \delta t = \sqrt{n} \delta T \propto n [/itex]

So, [itex] \delta T \propto \sqrt{n} [/itex]

Therefore, measuring the least number of cycle is best

Is it correct?