Answer
Both functions have the same amplitudes and periods but the graph of function $g$ is a vertical shift of $3$ units up of function $f$.
Work Step by Step
Using the equation of sine $y=a\sin b(x-c)+d$:
For $f(x)=\sin x=1\sin 1(x-0)+0$,
$a=1$
$period=\frac{2\pi}{b}=\frac{2\pi}{1}=2\pi$
$horizontal~shift=c=0=none$
$vertical~shift=d=0=none$
For $f(x)=3+\sin 2x=1\sin 2(x-0)+3$,
$a=1$
$period=\frac{2\pi}{b}=\frac{2\pi}{1}=2\pi$
$horizontal~shift=c=0=none$
$vertical~shift=d=3$
Thus, both functions have the same amplitudes and periods but the graph of function $g$ is a vertical shift of $3$ units up of function $f$.
