# I'm getting two possible uncertainties for frequency

• Andrew Tom
In summary: Sort of. You would do ##T\pm \Delta T = T\sqrt{(\frac{\Delta \lambda}{\lambda})^2+(\frac{\Delta \lambda}{\lambda})^2+(\frac{\Delta f}{f})^2}##.

#### Andrew Tom

Homework Statement
Find the uncertainty in the frequency of a wave
Relevant Equations
##v=\lambda f##
Calculate the speed and uncertainty in the speed for a wave with wavelength ##50\pm 3\mu s## and frequency ##30\pm 1kHz##. Also calculate the period and uncertainty in the period.

I am not sure about my answer to the second part. I used propagation of errors to get ##\Delta T=\Delta(1/f)=(\frac{1}{f})\sqrt{(\frac{\Delta 1}{1})^2+(\frac{\Delta f}{f})^2}=\frac{1}{f}(\frac{\Delta f}{f})=\frac{\Delta f}{f^2}=1.1111\mu s##. So ##T\pm \Delta T = 33.333\mu s \pm 1.111\mu s = 33\pm 1 \mu s##.

The reason I am not sure is because I did ##\Delta 1/1## which I have not seen before (but I only started learning uncertainties). Also the first part of the question is to calculate speed and its uncertainty, and if I use that to find uncertainty in period I get a different answer:

Using propagation of errors again with ##v=\lambda f## gives ##v\pm \Delta v = 1.5\pm 0.1029... =1.5\pm 0.1ms^{-1}##.

Then we can do ##\Delta T = \Delta (\frac{\lambda}{v})=\frac{1}{f}\sqrt{(\frac{\Delta \lambda}{\lambda})^2+\frac{\Delta v}{v})^2}=2.989...\mu s##. I.e. ##T\pm \Delta T = 33.3333\mu s \pm 2.989...\mu s = 33\pm 3\mu s##. This is different from the answer I got above without using ##v=\lambda f##, so I am a bit confused why there are different formulas using propagation of errors formula.

Also is there an easier way of doing this? Someone told me that the percentage error in ##f## would actually be the same as that in ##T## but I am not so sure.

Andrew Tom said:
Homework Statement:: Find the uncertainty in the frequency of a wave
Relevant Equations:: ##v=\lambda f##

Calculate the speed and uncertainty in the speed for a wave with wavelength ##50\pm 3\mu s## and frequency ##30\pm 1kHz##. Also calculate the period and uncertainty in the period.

I am not sure about my answer to the second part. I used propagation of errors to get ##\Delta T=\Delta(1/f)=(\frac{1}{f})\sqrt{(\frac{\Delta 1}{1})^2+(\frac{\Delta f}{f})^2}=\frac{1}{f}(\frac{\Delta f}{f})=\frac{\Delta f}{f^2}=1.1111\mu s##. So ##T\pm \Delta T = 33.333\mu s \pm 1.111\mu s = 33\pm 1 \mu s##.

The reason I am not sure is because I did ##\Delta 1/1## which I have not seen before (but I only started learning uncertainties). Also the first part of the question is to calculate speed and its uncertainty, and if I use that to find uncertainty in period I get a different answer:

Using propagation of errors again with ##v=\lambda f## gives ##v\pm \Delta v = 1.5\pm 0.1029... =1.5\pm 0.1ms^{-1}##.

Then we can do ##\Delta T = \Delta (\frac{\lambda}{v})=\frac{1}{f}\sqrt{(\frac{\Delta \lambda}{\lambda})^2+\frac{\Delta v}{v})^2}=2.989...\mu s##. I.e. ##T\pm \Delta T = 33.3333\mu s \pm 2.989...\mu s = 33\pm 3\mu s##. This is different from the answer I got above without using ##v=\lambda f##, so I am a bit confused why there are different formulas using propagation of errors formula.

Also is there an easier way of doing this? Someone told me that the percentage error in ##f## would actually be the same as that in ##T## but I am not so sure.
I didn't calculate the percent error but in your first propagation of errors it's ##\Delta \lambda = 3 ~ \mu s## and ##\Delta f = 1 ~ kHz##. You dropped the wavelength uncertainty for some reason.

-Dan

Also is there an
topsquark said:
I didn't calculate the percent error but in your first propagation of errors it's ##\Delta \lambda = 3 ~ \mu s## and ##\Delta f = 1 ~ kHz##. You dropped the wavelength uncertainty for some reason.

-Dan
I didn't use the wavelength uncertainty in my first calculation. I just wrote T=1/f and used propagation formula for division, with 1 and f as the "variables". I dropped the first term under square root because I thought ##\frac{\Delta 1 }{1} =0##.

You have to calculate the uncertainty in the calculated dependent variable(s) using the uncertainty in the measured independent variable(s). Here, the measured independent variables are the frequency and the wavelength. Whatever uncertainty you calculate must have ##\frac{\Delta f}{f}## and ##\frac{\Delta \lambda}{\lambda}## only on the right-hand side. Putting the uncertainty of a dependent variable on the right-hand side compounds the uncertainty unnecessarily.

Also, ##\frac{\Delta 1}{1}## is a silly way to look at differences. Number ##1## is always ##1## even for small values of ##1##. The way to handle this is formally, using the meaning of ##\Delta## as a difference operator: $$\Delta T=\Delta \left(\frac{1}{f}\right)=\frac{1}{f+\Delta f}-\frac{1}{f}=\frac{f-f-\Delta f}{f(f+\Delta f)}=\frac{-\Delta f}{f(f+\Delta f)}\approx -\frac{\Delta f}{f^2}.$$The last approximation is valid in the limit ##\Delta f \rightarrow 0.##

topsquark
Andrew Tom said:
I dropped the first term under square root because I thought ##\frac{\Delta 1 }{1} =0##.
You thought correctly.

kuruman said:
You have to calculate the uncertainty in the calculated dependent variable(s) using the uncertainty in the measured independent variable(s). Here, the measured independent variables are the frequency and the wavelength. Whatever uncertainty you calculate must have ##\frac{\Delta f}{f}## and ##\frac{\Delta \lambda}{\lambda}## only on the right-hand side. Putting the uncertainty of a dependent variable on the right-hand side compounds the uncertainty unnecessarily.
So would I do ##\Delta T = \Delta \frac{\lambda}{\lambda f}=T\sqrt{(\frac{\Delta \lambda}{\lambda})^2+(\frac{\Delta \lambda}{\lambda})^2+(\frac{\Delta f}{f})^2}##? Somehow this doesn't seem right.

kuruman said:
Also, ##\frac{\Delta 1}{1}## is a silly way to look at differences. Number ##1## is always ##1## even for small values of ##1##. The way to handle this is formally, using the meaning of ##\Delta## as a difference operator: $$\Delta T=\Delta \left(\frac{1}{f}\right)=\frac{1}{f+\Delta f}-\frac{1}{f}=\frac{f-f-\Delta f}{f(f+\Delta f)}=\frac{-\Delta f}{f(f+\Delta f)}\approx -\frac{\Delta f}{f^2}.$$The last approximation is valid in the limit ##\Delta f \rightarrow 0.##
Ah ok, so would differentiation be better? I.e. ##\Delta T=\Delta \frac{1}{f}=\frac{dT}{df}\Delta f = -\frac{\Delta f}{f^2}##?

Andrew Tom said:
Ah ok, so would differentiation be better? I.e. ##\Delta T=\Delta \frac{1}{f}=\frac{dT}{df}\Delta f = -\frac{\Delta f}{f^2}##?
Yes, taking derivatives would be quicker if you are familiar with them. The derivation in post #4 is really taking the derivative using the formal definition $$\frac{dT}{df}=\lim_{\Delta f \rightarrow 0} {\frac {T(f+\Delta f)-T(f)}{\Delta f}}.$$

topsquark
Andrew Tom said:
Ah ok, so would differentiation be better? I.e. ##\Delta T=\Delta \frac{1}{f}=\frac{dT}{df}\Delta f = -\frac{\Delta f}{f^2}##?
I guess that's one way to look at it.

## What does it mean to have two possible uncertainties for frequency?

Having two possible uncertainties for frequency means that there are two different values that could represent the true value of the frequency. This can happen when there is a margin of error in the measurement or when there is a range of values that could be considered acceptable.

## How do you calculate the uncertainties for frequency?

The uncertainties for frequency can be calculated using the formula: Δf = fmax - fmin, where fmax is the highest possible frequency and fmin is the lowest possible frequency. This will give you a range of values that could represent the true frequency.

## What factors can contribute to having two possible uncertainties for frequency?

There are several factors that can contribute to having two possible uncertainties for frequency. These can include limitations in the measuring equipment, external influences such as noise or interference, and human error in recording the data.

## How do you determine which uncertainty value to use?

The uncertainty value to use will depend on the level of accuracy required for the specific application. If a higher level of precision is needed, it may be necessary to use the smaller uncertainty value. However, if the accuracy requirement is lower, the larger uncertainty value may be sufficient.

## What is the significance of having two possible uncertainties for frequency in scientific research?

In scientific research, having two possible uncertainties for frequency can indicate that there is a level of uncertainty in the data being collected. This can be important to consider when drawing conclusions or making decisions based on the data. It is also important to report both uncertainties to provide a complete understanding of the data.

• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
605
• Introductory Physics Homework Help
Replies
10
Views
775
• Introductory Physics Homework Help
Replies
4
Views
952
• Introductory Physics Homework Help
Replies
19
Views
587
• Introductory Physics Homework Help
Replies
1
Views
639
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
917
• Introductory Physics Homework Help
Replies
25
Views
337
• Introductory Physics Homework Help
Replies
9
Views
1K