Phase difference between 2 points on a wave

[Bolter]

Homework Statement

Calculate the phase difference

Relevant Equations

See below

So to do this problem I need the relevant formula for phase difference which is this:

I first need to find wavelength and this is lambda = velocity/frequency

So lambda = 257/641 = 0.40093603744 m
Hence phase difference (in radians) = 2pi * (2/0.40093603744) = 31.3 rads

My concern is that doesn't the phase difference always have to be less than 2pi radians? i.e. the path difference must always be less than the wavelength?

I know that exactly 4 wavelengths fit exactly into 2 meters and then you have 0.39625585 m left over. So should 0.39625585m be the path difference, thus the phase difference then becomes

phase difference = 2pi * (0.39625585/0.40093603744) = 6.2 radians

Any help would be really great! Thanks a lot

 

[PeroK]

I would say that 257/641=0.401. Assuming we have the data to three significant figures.

 

[Cutter Ketch]

My concern is that doesn't the phase difference always have to be less than 2pi radians?

Yes, that is correct. However, it is perfectly reasonable to get more than 2pi. You can go around a circle many times and still answer the question “how far have you moved around the circle from where you started?” You just have to subtract all the complete circles. Subtract the largest possible multiple of 2pi, and that’s the phase.

Now, knowing that, people sometimes casually use the word “phase” without subtracting. “Passing through 4 mm of glass the wave accumulates 11,000 waves of phase.” But when you are asked what is “the phase” or “the phase difference” the answer should be the proper definition: the extra part after subtracting any complete waves.

 

[Cutter Ketch]

P.S.: since the phase is the residual after subtracting what sometimes is a large number of waves, the calculation needs high precision. Don’t round off anything until you get to the final answer.

 

[Bolter]

Yes, that is correct. However, it is perfectly reasonable to get more than 2pi. You can go around a circle many times and still answer the question “how far have you moved around the circle from where you started?” You just have to subtract all the complete circles. Subtract the largest possible multiple of 2pi, and that’s the phase.

Now, knowing that, people sometimes casually use the word “phase” without subtracting. “Passing through 4 mm of glass the wave accumulates 11,000 waves of phase.” But when you are asked what is “the phase” or “the phase difference” the answer should be the proper definition: the extra part after subtracting any complete waves.

So either one of 31.3 or 6.2 rads is acceptable for the phase difference then. But of course 6.2 rad is a more suitable value as it’s line with taking into account the extra part of the wave after subtracting all number of complete waves. It corresponds to what the definition of phase difference is

 

[Bolter]

P.S.: since the phase is the residual after subtracting what sometimes is a large number of waves, the calculation needs high precision. Don’t round off anything until you get to the final answer.

I have kept everything to a high precision when entering it onto the calculator. I rounded both my values to 1 decimal place to get 31.3 and 6.2 rads

 

[haruspex]

I take a different view.
There is no point stating a phase difference with a magnitude greater than 2 pi. The number of wavelengths is (2m)(641Hz)/(257m/s). Take the difference from the nearest whole number and then multiply by 2π.

 

[Cutter Ketch]

I take a different view.
There is no point stating a phase difference with a magnitude greater than 2 pi. The number of wavelengths is (2m)(641Hz)/(257m/s). Take the difference from the nearest whole number and then multiply by 2π.

Different view? I believe that is consistent with what I said. I said people sometimes casually use the word “phase” to refer to the whole enchilada, but the proper definition is just the angle within 2pi.

 

[haruspex]

Different view?

From that in posts #5 and #6.
But also, by working in wavelengths first, it is not necessary to plug in the value of pi to high precision - not that that is at all arduous given that it will be supplied to great precision by the calculator.