In physics, the wave function in its various forms (Serway 503-504) is given by \begin{align} y&=A\sin{\left[\frac{2\pi}{\lambda}\left(x-vt\right)\right]}\\ &=A\sin{\left[2\pi\left(\frac{x}{\lambda}-\frac{t}{T}\right)\right]}\\ &=A\sin{\left(kx-\omega t\right)} \vphantom{\left[\left(\frac{x}{\lambda}-\frac{t}{T}\right)\right]}\\ \end{align} In trigonometry, the function $y$ given by \begin{align} y&=A\sin{B\left(x-\frac{C}{B}\right)}\\ &=A\sin{\left(Bx-C\right)} \vphantom{\left(x-\frac{C}{B}\right)} \end{align} has period $\frac{2\pi}{B}$ and phase shift $\frac{C}{B}$ (Rice 197). Recall that the fundamental period of $y$ is the smallest positive number $p$ satisfying \begin{equation} y\left(x\right)=y\left(x+p\right) \end{equation} for all $x$ (Rice 175), and the phase shift is the value of $x$ that makes the argument of the sine or cosine function $0$ (Rice 197). Recall also that if $y$ is periodic with period $p,$ then it is also periodic with any integral multiple of $p$ (Rice 175). That is, \begin{equation} y\left(x\right)=y\left(x+n\lambda\right) \end{equation} for any integer $n.$

To apply the preceding knowledge to the wave function, we first note that since $y$ is a function of $x$ and $t,$ then for constant $t$ and variable $x,$ we will obtain one period and phase shift, and for constant $x$ and variable $t$ another period and phase shift. Note, however, that in the context of physics, when we speak of the period of the wave function, we are speaking in particular of the period $T$ when $x$ is held constant, the time it takes for the wave to travel a distance of one wavelength (Serway 492). In the general trigonometric sense, the case of holding $t$ constant, gives the period as, \begin{equation} \frac{2\pi}{2\pi/\lambda}=\lambda \end{equation} implying, \begin{equation} y\left(x,t_0\right)=y\left(x+n\lambda,t_0\right) \end{equation} Therefore, any multiple of the wavelength produces the same value of $y,$ vertical displacement. The phase shift is \begin{equation} \frac{t_0}{T} \end{equation} implying that the phase shift is $0$ when $t_0=0.$ If a phase shift is desired even when $t_0=0,$ we add the phase constant $\phi$ to the argument of the sine function in the wave function to obtain the new wave equation \begin{align} y&=A\sin{ \left[ \frac{2\pi}{\lambda} \left(x-vt+\frac{\lambda\phi}{2\pi}\right) \right]}\\ &=A\sin{ \left[ 2\pi \left( \frac{x}{\lambda} -\frac{t}{T} +\frac{\phi}{2\pi} \right) \right]}\\ &=A\sin{\left(kx-\omega t+\phi\right)} \vphantom{ \left( \frac{x}{\lambda} -\frac{t}{T} +\frac{\phi}{2\pi} \right) } \end{align} From this we see that the phase shift is $vt_0-\frac{\lambda\phi}{2\pi}=vt_0-\frac{\phi}{k}.$ Thus, when $t_0=0,$ the phase shift is $\frac{\lambda\phi}{2\pi}=\frac{\phi}{k}.$ Note that $\phi$ is called the phase constant (Serway 505), not the phase shift; different things.

Repeating the analysis of the wave function now for constant $x$ and variable $t,$ we start by rearranging the wave function to put it in the general form: \begin{align} y&=A\sin{ \left[ \frac{-2\pi v}{\lambda} \left(t-\frac{x_0}{v}\right) \right]}\\ &=A\sin{ \left[ \frac{-2\pi}{T}\left(t-\frac{x_0}{v}\right) \right]}\\ &=A\sin{\left[-\left(\omega t-kx\right)\right]} \vphantom{ \left[ \frac{-2\pi}{T}\left(t-\frac{x_0}{v}\right) \right] } \end{align} The period, therefore, is \begin{equation} \frac{2\pi}{2\pi/T}=T \end{equation} so that \begin{equation} y\left(x_0,t\right)=y\left(x_0,t+nT\right) \end{equation} The phase shift is \begin{equation} \frac{x_0}{v} \end{equation} As before, if we wish a phase shift even when $x_0=0,$ we introduce a phase constant $\phi.$

Works Cited

Serway, Raymond A. and Robert J. Beichner. Physics for Scientists and Engineers, Fifth Edition. Brooks/Cole, 2000. Print.

Rice, Bernard J. and Jerry D. Strange. Plane Trigonometry, Seventh Edition. Boston: PWS Publishing Company, 1996. Print