# I'm getting two possible uncertainties for frequency

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.

Gold Member
MHB
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

Andrew Tom
Also is there an
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##.

Homework Helper
Gold Member
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
Homework Helper
Gold Member
I dropped the first term under square root because I thought ##\frac{\Delta 1 }{1} =0##.
You thought correctly.

Andrew Tom
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.

Andrew Tom
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}##?

Homework Helper
Gold Member
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