Phase Shift

Sample sine sinusoidal as a function of x with an angular wave number and phase constant.

The determination of the phase constant is not unknowable magic. It is the result of some simple algebra. Let's look at the general form for a sinusoidal equation without the complication of a vertical shift.

$$y=A\sin \left(kx+{\phi }_0\right)$$

We can use the horizontal shift rule to predict the sign of the phase constant:

“If the argument x of a function f is replaced by x − h , h > 0 , the graph of the new function y = f ( x − h ) is the graph of f shifted horizontally right h units. If the argument x of a function f is replaced by x + h , h > 0 , the graph of the new function y = f ( x + h ) is the graph of f shifted horizontally left h units.”

— Sullivan and Sullivan, p. 246 [1]

From inspecting the figure above, we can tell that the amplitude, A, of the oscillation is 10 cm, and at x=0, the displacement, y = 5. From the horizontal shift rule, we expect the phase constant to be positive.

$$\begin{array}{c}5=10\sin \left(k\cdot 0+{\phi }_0\right)\\ \frac{1}{2}=\sin \left({\phi }_0\right)\\ {\sin }^{-1}\left(\frac{1}{2}\right)={\phi }_0\\ {\phi }_0=\frac{\pi }{6}\text{ or }30°\end{array}$$

The phase constant is positive, as expected.

We can also calculate the angular wave number, k, by choosing another point on the figure. Let's choose x=—5. y=10: 

$$\begin{array}{c}10=10\sin \left(k\cdot -5+\frac{\pi }{6}\right)\\ 1=\sin \left(-5k+\frac{\pi }{6}\right)\\ {\sin }^{-1}\left(1\right)=-5k+\frac{\pi }{6}=-\frac{3\pi }{2}\\ -5k=-\frac{3\pi }{2}-\frac{\pi }{6}=-\left(\frac{18\pi }{12}-\frac{2\pi }{12}\right)\\ k=\frac{\pi }{5}\left(\frac{20}{12}\right)=\frac{2\pi }{6}\\ k=\frac{\pi }{3}\end{array}$$

We have the complete displacement equation:

$$y=10\sin \left(\frac{\pi }{3}x+\frac{\pi }{6}\right)$$

The sample sine sinusoidal as a function of theta.

We can check our work by evaluation displacement function a other points on the figure:

$$\begin{array}{l}y=10\sin \left(\frac{\pi }{3}10+\frac{\pi }{6}\right)=-10\\ y=10\sin \left(\frac{\pi }{3}20+\frac{\pi }{6}\right)=5\end{array}$$

They match, we can be confident in our answer.

We could also look at this as a change of variable problem.

What would the solution be if the sinusoidal function was a cosine function instead of the sine function?

Note: The arcsine function is a multi-value function. Be sure to understand why we choose 

$${\sin }^{-1}\left(1\right)=-\frac{3\pi }{2}$$

instead of one of the infinite other possibilities.

[1] Sullivan, Michael, and Michael Sullivan (III). 2012. “Graphing Techniques: Graph Functions Using Vertical and Horizontal Shifts.” In Algebra & Trigonometry Enhanced with Graphing Utilities, 6th ed., 244–47. Boston: Pearson Education. http://bit.ly/1OxChzf.