Answer
Amplitude is $3$ and the period is $4\pi $
Work Step by Step
The given equation is $y=3\sin \frac{1}{2}x$. It is in the form of $y=A\operatorname{sinBx}$, A is 3 and B is $\frac{1}{2}$. It shows the amplitude of 3 and the period is,
$\begin{align}
& \frac{2\pi }{B}=\frac{2\pi }{\frac{1}{2}} \\
& =2\pi \times 2 \\
& =4\pi
\end{align}$
The quarter period is,
$\frac{4\pi }{4}=\pi $
Since it is mentioned that the cycles begin from 0, we add the quarter period in order to generate the values of x. Find the various values of the given function as follows:
When the value of x is 0:
$\begin{align}
& y=3\sin \left( \frac{1}{2}x \right) \\
& =3\sin \left( \frac{1}{2}\times 0 \right) \\
& =0
\end{align}$
And the coordinates of the graph are:
$\left( 0,0 \right)$.
When the value of x is $\pi $:
$\begin{align}
& y=3\sin \left( \frac{1}{2}x \right) \\
& =3\sin \left( \frac{1}{2}\times \pi \right) \\
& =3\sin \left( \frac{\pi }{2} \right)
\end{align}$
And the value at $\frac{\pi }{2}$ is 1. So,
$\begin{align}
& y=3\times 1 \\
& =3
\end{align}$
And the coordinates of the graph are $\left( \pi,3 \right)$.
When the value of x is $2\pi $:
$\begin{align}
& y=3\sin \left( \frac{1}{2}x \right) \\
& =3\sin \left( \frac{1}{2}\times 2\pi \right) \\
& =3\sin \left( \frac{2\pi }{2} \right)
\end{align}$
The value at $\pi $ is 0. Then,
$\begin{align}
& y=3\times 0 \\
& =0
\end{align}$
And the coordinates of the graph are $\left( 2\pi,0 \right)$.
When the value of x is $3\pi $. So,
$\begin{align}
& y=3\sin \left( \frac{1}{2}x \right) \\
& =3\sin \left( \frac{1}{2}\times 3\pi \right) \\
& =3\sin \left( \frac{3\pi }{2} \right)
\end{align}$
The value at $\frac{3\pi }{2}$ is -1
$\begin{align}
& y=3\times -1 \\
& =-3
\end{align}$
And the coordinates of the graph are $\left( 3\pi,-3 \right)$.
When the value of x is $4\pi $. So,
$\begin{align}
& y=3\sin \left( \frac{1}{2}x \right) \\
& =3\sin \left( \frac{1}{2}\times 4\pi \right) \\
& =3\sin \left( \frac{4\pi }{2} \right)
\end{align}$
The value of $2\pi $ is 0,
$\begin{align}
& y=3\times 0 \\
& =0
\end{align}$
