- #1

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- 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 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.