I am currently stuck on a problem in my physics book but its the math that I am having trouble with. I am trying to find the phase shift of a sine function of the form $$ s(x,t) = A\,\sin \left\{ 2\pi \left(\frac{t}{T} \pm \frac{x}{W}\right) + \phi \right\} .$$ Where $W=6$, $T=22$ and $A=0.15$.
My work:
Phase shift $\phi$: At the point t=0 and x=2 the function has a value of 0, this should imply that $\sin\{( 2\pi(-\frac{2}{6})+ \phi\}=0$, since the term with t would become 0 and $\sin^{-1}(0) = \pm \pi$ this means that $2\pi(\frac{-2}{6}) + \phi =\pm\pi$ and we should get two phase-shifts. 5π/3 and $-\pi/3$. This is correct. If I try using the same logic with an extreme point such as $x=3.5$ I also get the same phase shifts. But if i try with a point like $x=5$ I do not get the correct phase shifts. I end up with $8\pi/3$ and $-4\pi/3$ which is not correct. Why is this?
I frankly do not understand how your work is unfolding. Just read your graphs. Your left graph is $$ s(x,0)=A \sin\left (\phi_{\pm}\pm 2\pi( \frac{x}{6})\right ), $$ for left, and right-moving waves, respectively.
For the left-moving wave, your function values 0,A,0, specify x = 2, 3.5, 5, respectively, so confirm that $\phi_+ = -2\pi/3$. Confirm this is in good agreement with the right graph!
For the right-moving wave, instead, your x values are in descending order of the fundamental cycle, so the same values of the function are at x=5, 3.5, 2. Consequently, $\phi_-= 5\pi/3$.
Observe the phases of the right and left moving waves add up to half a cycle, π.
Your error appears to be the expectation that $\phi_+=\phi_-$.
You might interpret your phase shifts as follows. For the left-moving wave, +, $\phi_+$ is the amount you need to shift the true origin of the cycle, x=2, to the left, to position it at the crossing of the axes, in your left graph. For the right-moving wave, $-$, you need to do the same to x=5, by $-\phi_-$ to get minus the standard cycle, as dictated by the above sine function.