# How do you find the amplitude, period, and phase shift for a sine function?

Nov 5, 2015

If you have $A \sin \left(\omega x + \phi\right)$, then:

• $A$ is the amplitude;
• $\frac{2 \pi}{\omega}$ is the period;
• $\phi$ is the shift.

#### Explanation:

The most general case of a sine function is

$A \sin \left(\omega x + \phi\right)$. Let's see what they do.

• Since the sine function is bounded between $- 1$ and $1$, if we multiply the function by $A$ the result will be bounded between $- A$ and $A$. So, the amplitude is "decided" by the outter coefficient.

• The $\omega$ coefficient affects the period: we know that "normally" $x$ runs a full period from $0$ to $2 \pi$. Multipling $x$ by a factor $w$ means to change the speed of the forementioned run. For example, if you change your variable from $x$ to $2 x$, your variable will be running at twice the speed, and so you will go from $0$ to $2 \pi$ as $x$ goes from $0$ to $\pi$.
On the other hand, if you change from $x$ to $\frac{x}{2}$, you will go at half the speed, and so $\frac{x}{2}$ goes from $0$ to $2 \pi$ (i.e, runs a full period) as $x$ goes from $0$ to $4 \pi$.
So, we see that the period is $\frac{2 \pi}{\omega}$

• Finally, the term $\phi$ is the shift, it simply means that instead of starting from zero, you start from $\phi$: take for example $\sin \left(x\right)$ and $\sin \left(x + \frac{\pi}{3}\right)$. When $x$ is zero, $\sin \left(x + \frac{\pi}{3}\right)$ is "already" at $\frac{\pi}{3}$, representing what I said before: the function starts from $\phi$ at not from zero.