Uncertainty of an oscilloscope frequency, read from the period

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Homework Help Overview

The discussion revolves around calculating the uncertainty of frequency derived from the period measured using a cathode ray oscilloscope (CRO). Participants are exploring the relationship between period and frequency, particularly focusing on how to propagate uncertainty in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of calculating uncertainty in frequency based on the uncertainty in period. There is a question about whether the relationship can be treated as a polynomial function and how to apply error propagation correctly.

Discussion Status

The conversation is active, with participants providing guidance on the use of derivatives for uncertainty propagation. There is a recognition that the percent uncertainty in frequency is related to the percent uncertainty in period, which some participants find helpful for understanding the problem.

Contextual Notes

Some participants are questioning the assumptions regarding the treatment of the function and the values used for calculations, indicating a need for clarity on the definitions and relationships involved.

zzzriprip
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Homework Statement
Uncertainty for lab reports
Relevant Equations
T = 1/f
Hi, I am unsure of what uncertainty to get, so here is my full question: I used the CRO for an experiment, and since what I need is frequency, I read the period, so for the uncertainty of the period, it is the smallest division divided by two. So if my uncertainty for period is 0.001s, then what would be the uncertainty of the frequency? Do I need to use error propagation?
 
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Yes you do, but that is always true. So how do you approach this one?
 
hutchphd said:
Yes you do, but that is always true. So how do you approach this one?
Is it considered as a polynomial function? since f = 1/T, so f = t^-`1, thus df/dt = -1/t^2, and then do I just substitute t=0.001 into the equation to get the uncertainty?
 
zzzriprip said:
thus df/dt = -1/t^2
Right.
zzzriprip said:
do I just substitute t=0.001 into the equation to get the uncertainty?
No.
##\Delta f=\frac{df}{dt}\Delta t##.
The derivative needs to be evaluated using the value of t measured.
 
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zzzriprip said:
Is it considered as a polynomial function? since f = 1/T, so f = t^-`1, thus df/dt = -1/t^2, and then do I just substitute t=0.001 into the equation to get the uncertainty?
As @haruspex points out the 0.001s is Δt and you need to also put in t. If you do a little algebra on your result above you can show for this case the percent uncertainty in f will be the same as the percent uncertainty in t (i.e. Δt/t). That makes it easy to understand and use.
 
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