y = A sin(ωx + φ): amplitude, period and phase shift

y = A sin(ωx + φ)

What do A, ω and φ each do to the graph of y = sin x?

Three sliders turn y = sin x into y = A sin(ωx + φ). Watch the amplitude, the period 2π/ω and the starting point −φ/ω move, and see why sin(2x + π/3) shifts by π/6, not π/3.

High school
Amplitude
Period T = 2π/ω
Phase shift
Order of transformations

A

2.0

\omega

2.0

\varphi\,(^\circ)

y = A\sin(\omega x + \varphi)

$A$: 2.0
$T = \dfrac{2\pi}{\omega}$: 3.142
$\varphi\ (\text{rad})$: 1.047
$-\dfrac{\varphi}{\omega}$: -0.524
$x_{\text{peak}}$: 0.262
$f(0) = A\sin\varphi$: 1.732

Guided tour

1Start from y = sin x
With $A=1$ , $\omega=1$ , $\varphi=0$ the curve is $y=\sin x$ : it starts rising at the origin, peaks at $1$ and repeats every $T=2\pi\approx6.283$ .
2A sets the height
At $A=2$ every value doubles: the curve now runs between $-2$ and $2$ . The zeros and the period stay where they were, because $A$ only acts on $y$ .
3ω sets the period
At $\omega=2$ the angle $2x$ runs twice as fast, so one full wave fits into $T=\frac{2\pi}{2}=\pi\approx3.142$ . Larger $\omega$ squeezes the graph horizontally.
4φ shifts by φ/ω, not by φ
Set $\varphi=60^\circ=\frac{\pi}{3}$ . The wave starts where $2x+\frac{\pi}{3}=0$ , at $x=-\frac{\pi}{6}\approx-0.524$ : it moved **left by $\frac{\pi}{6}$ **. Write it as $y=2\sin 2\left(x+\frac{\pi}{6}\right)$ to see why. The first peak is at $x=\frac{\pi}{12}\approx0.262$ .

Explanation

1
Three parameters, three jobs
$y=A\sin(\omega x+\varphi)\qquad(A>0,\ \omega>0)$
- Amplitude $A$ : the curve runs between $-A$ and $A$ .
- Period $T=\dfrac{2\pi}{\omega}$ : the length of one full wave. Its reciprocal $f=\dfrac1T$ is the frequency.
- Initial phase $\varphi$ : the phase $\omega x+\varphi$ at $x=0$ , so $f(0)=A\sin\varphi$ .
2
Finding the shift
Factor $\omega$ out of the bracket:
$A\sin(\omega x+\varphi)=A\sin\omega\left(x+\frac{\varphi}{\omega}\right)$
So the graph of $y=A\sin\omega x$ moves by $\dfrac{\varphi}{\omega}$ : left when $\varphi>0$ , right when $\varphi<0$ . A reliable shortcut is to solve $\omega x+\varphi=0$ for the starting point of the wave.
3
Two orders, two different shifts
From $y=\sin x$ to $y=\sin\left(2x+\frac{\pi}{3}\right)$ :
- Shift first: left by $\frac{\pi}{3}$ to get $\sin\left(x+\frac{\pi}{3}\right)$ , then halve every $x$ -coordinate.
- Stretch first: halve every $x$ -coordinate to get $\sin 2x$ , then shift left by only $\frac{\pi}{6}$ .
Both arrive at the same curve. The shift is different because in the second route it acts on $x$ after $x$ has already been scaled by $\omega$ .
4
Peaks, zeros and the five key points
Set the phase $\omega x+\varphi$ to $0,\ \frac{\pi}{2},\ \pi,\ \frac{3\pi}{2},\ 2\pi$ and solve for $x$ . These five points (zero, peak, zero, trough, zero) are a quarter period apart and are all you need to sketch one period by hand. The peak value is $A$ , reached where $\omega x+\varphi=\frac{\pi}{2}+2k\pi$ .

Check yourself

Q01

What is the period of $y=3\sin\left(4x-\frac{\pi}{5}\right)$ ?

Q02

How is $y=\sin\left(2x-\frac{\pi}{3}\right)$ obtained from $y=\sin 2x$ ?

Q03

What is the maximum value of $y=2\sin\left(x+\frac{\pi}{4}\right)$ ?

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