GIVEN AMPLITUDE, PERIOD, and PHASE SHIFT, WRITE AN EQUATION
PRACTICE (online exercises and printable worksheets)
In the
prior section,
you learned how to find the amplitude, period, and phase shift of a given
(generalized) sine or cosine curve.
In this section, you will write an
equation of a curve with a specified amplitude, period, and phase shift.
Sample question:
Write an equation of a sine curve with amplitude $\,5\,$, period $\,3\,$, and phase shift $\,2\,$.
Specifying a sine (or cosine) curve with a given amplitude, period, and phase shift
defines a unique set of points in the plane.
However, there are infinitely many equations
that can describe that set of points!
For example, the set of points given by the sine curve with amplitude $\,5\,$, period $\,3\,$, and phase shift $\,2\,$ can be described
by any of these equations (and many more):
$\displaystyle y = 5\sin \frac{2\pi}{3}(x - 2)$
$\displaystyle y = -5\sin \frac{2\pi}{3}(x - 3.5)$
$\displaystyle y = 5\sin \frac{2\pi}{3}(x + 1)$
In this section, we find a simple (natural) equation that works,
by using graphical transformations to change
a basic model into a curve with the desired attributes.
The process is illustrated with an example:
EXAMPLE:
Write an equation of a sine curve with amplitude $\,5\,$, period $\,3\,$, and phase shift $\,2\,$.
SOLUTION:
Start with the basic model (sine or cosine):
We want a sine curve, so the ‘basic model’ is: $y = \sin x$
Apply a vertical stretch/shrink to get the desired amplitude:
new equation: $y = 5\sin x$
Apply a horizontal stretch/shrink to get the desired period:
For $\,k > 0\,$, the curve $\,y = \sin kx\,$ has period $\displaystyle\,\frac{2\pi}{k}\,$.
The basic sine curve, $\,y = \sin t\,$, goes through one complete cycle as its inputs ($\,t\,$) go from $\,\color{purple}{0}\,$ to $\,\color{green}{2\pi}\,$.
So, how long does it take for $\,y = \sin kx\,$ to go through one complete cycle?
The inputs for $\,y = \sin \color{blue}{kx}\,$ are $\,\color{blue}{kx}\,$.
When are the inputs $\,\color{purple}{0}\,$?
$\color{blue}{kx} = \color{purple}{0}$
when
$\color{red}{x = 0}$
When are the inputs $\,\color{green}{2\pi}\,$?
$\color{blue}{kx} = \color{green}{2\pi}$
when
$\displaystyle\color{red}{x = \frac{2\pi}k}$
Thus, $\,y =\sin kx\,$ goes through one complete cycle as $\,x\,$ goes from $\,\color{red}{0}\,$ to $\displaystyle\,\color{red}{\frac{2\pi}k}\,$.
Thus, the period of $\,y =\sin kx\,$ is: $\,\displaystyle\color{red}{\frac{2\pi}k} - \color{red}{0} = \frac{2\pi}k\,$
We want the period to be $\,3\,$:
$\displaystyle \frac{2\pi}{k} = 3\ \ \implies\ \ k = \frac{2\pi}{3}$
new equation: $\displaystyle\,y = 5\sin \frac{2\pi}3x\,$
Apply the desired phase shift:
To shift right $\,2\,$, replace every $\,x\,$ by $\,x - 2\,$.
Remember—transformations involving $\,x\,$ are counter-intuitive!
new equation: $\displaystyle y = 5\sin \frac{2\pi}{3}(x - 2)$
Thus, the curve $\displaystyle y = 5\sin \frac{2\pi}{3}(x - 2)$ has amplitude $\,5\,$, period $\,3\,$, and phase shift $\,2\,$.
Important: Change the period BEFORE the phase shift!
If you apply the phase shift first, then the subsequent horizontal stretch/shrink to adjust the
period will mess up this phase shift.
So, be sure to adjust the period before applying the phase shift!
The amplitude adjustment (the vertical stretch/shrink transformation) can be applied at any time.
Here is a second example:
EXAMPLE:
Write an equation of a cosine curve with amplitude $\,4\,$, period $\displaystyle\,\frac{\pi}{3}\,$, and phase shift $\displaystyle\,-\frac 12\,$.
SOLUTION:
basic model: $y = \cos x$
period: $\displaystyle\frac{2\pi}{k} = \frac{\pi}{3}\ \ \implies\ \ k = 6$
new equation: $\displaystyle\,y = \cos 6x$
Thus, the curve $\displaystyle\,y = 4\cos 6(x+\frac 12)$ has amplitude $\,4\,$, period $\displaystyle\,\frac{\pi}{3}\,$, and phase shift $\displaystyle\,-\frac 12\,$.
Master the ideas from this section
by practicing the exercise at the bottom of this page.