Chapter 6, Problem 53RE

To determine

To find: Period , amplitude , phase shift and graph

Answer to Problem 53RE

(a) period $=\frac{2\pi }{B}=\frac{2\pi }{30\pi }=\frac{1}{15}$

  $\begin{array}{l}\text{(b)Aplitude}=220\\ \text{(c)phaseshift=}\frac{1}{180}\\ \text{(d)flow the above picture}\text{.}\end{array}$

Explanation of Solution

Given information: $I(t)=220\sin (30\pi t+\frac{\pi }{6})t\ge 0\mathrm{......}\text{(1)}$

$I(t)=A\sin (B(t+c))+D\mathrm{......}\text{(2)}$

Where, Amplitude is A

Period is $\frac{2\pi }{B}$

Phase shift is c (positive is to the left)

  $\frac{2\pi }{B}$

Amplitude

(vertical shift)

(Phase shift left)

Comparing (1) by (2)

Then $A=220,B=30\pi ,C\frac{1}{180},D=0$

Hence, (a) period $=\frac{2\pi }{B}=\frac{2\pi }{30\pi }=\frac{1}{15}$

  $\begin{array}{l}\text{(b)Aplitude}=220\\ \text{(c)phaseshift=}\frac{1}{180}\\ \text{(d)flow the above picture}\text{.}\end{array}$