Chapter 6.6, Problem 27AYU

To determine

To find: The period, amplitude, phase shift and the graph of the function.

Answer to Problem 27AYU

The period, amplitude, and the phase shift of the function is $\frac{1}{15}$ , $120$ , and $\frac{1}{90}$ .

Explanation of Solution

Given information:

The given function is,

  $I\left(t\right)=120\sin \left(30\pi t-\frac{\pi }{3}\right)$

The given function can be written as,

  $\begin{array}{c}I\left(t\right)=120\sin \left(30\pi t-\frac{\pi }{3}\right)\\ =120\sin \left(30\pi \left(t-\frac{1}{90}\right)\right)\end{array}$

The standard form of the equation is,

  $f\left(t\right)=A\sin \left(\omega \left(t-\frac{\varphi }{\omega }\right)\right)$

Compare the standard equation and the given function to obtain,

  $\begin{array}{l}A=120\\ \omega =30\pi \\ T=\frac{1}{15}\\ \frac{\varphi }{\omega }=\frac{1}{90}\end{array}$

Thus, the period, amplitude, and the phase shift of the function is $\frac{1}{15}$ , $120$ , and $\frac{1}{90}$ .

Consider the graph of the given function as follows,