Answer
$$\eqalign{
& {\text{amplitude: 2}}{\text{.5}} \cr
& {\text{period: 4}}\pi \cr} $$
Work Step by Step
$$\eqalign{
& p\left( t \right) = 2.5\sin \left( {\frac{1}{2}\left( {t - 3} \right)} \right) \cr
& {\text{The functions of the form }}y = A\sin \left( {B\left( {\theta - C} \right)} \right) + D{\text{ have a vertical}} \cr
& {\text{stretch }}\left( {{\text{or }}{\bf{amplitude}}} \right){\text{ of }}\left| A \right|,{\text{ and a period of }}\frac{{2\pi }}{{\left| B \right|}} \cr
& {\text{Therefore, comparing the given function we obtain}} \cr
& \underbrace {p\left( t \right) = 2.5\sin \left( {\frac{1}{2}\left( {t - 3} \right)} \right)}_{y = A\sin \left( {B\left( {\theta - C} \right)} \right) + D} \cr
& A = 2.5,{\text{ }}B = \frac{1}{2},{\text{ }}C = 3,{\text{ }}D = 0 \cr
& {\text{The amplitude is: }}\left| A \right| = 2.5 \cr
& {\text{Period: }}\frac{{2\pi }}{{\left| B \right|}} = \frac{{2\pi }}{{\left| {1/2} \right|}} = 4\pi \cr
& \cr
& {\text{amplitude: 2}}{\text{.5}} \cr
& {\text{period: 4}}\pi \cr} $$
