# Calculating phase difference of sound waves

An observer stands 3 m from speaker A and 5 m from speaker B. Both speakers, oscillating in phase, produce waves with a frequency of 250 Hz. The speed of sound in air is 340 m/s. What is the phase difference between the waves from A and B at the observer's location?

$$v=fλ\\ λ=1.36m\\ \\ Δr=|{ r }_{ 2 }-{ r }_{ 1 }|\\ =2$$

I have nothing that relates any of this to phase angle. :(

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Note that phase angle is not a real angle, it is just a convenient description of timing differences. One wavelength is $360^\circ$ by definition, so you simply find how large a fraction of a wavelength the waves are off and multiply that by $360^\circ$.

Depending on the context you might also want to normalize the result to the range $[0^\circ,180^\circ]$, that is first normalizing to the range $[-180^\circ,180^\circ]$ by adding or subtracting an integer number of $360^\circ$, then taking the absolute value.

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"you simply find how large a fraction of a wavelength the waves are off" They both have the same wavelength. So the waves are off by zero, right? –  user23276 Apr 17 '13 at 15:51
Nope. You have left out a unit in the numbers in your question, fixing that might help. –  eBusiness Apr 17 '13 at 16:00
I think I got it: 2m/1.36m*2pi = 9.24 rad = 2.96 rad Thank you! –  user23276 Apr 17 '13 at 16:46
Good. Always remember your units, you wrote 2 instead of 2m in your question, that might have confused you. –  eBusiness Apr 17 '13 at 16:55