# How do you find the amplitude and period of y=3sin(2t)?

##### 1 Answer
Nov 26, 2016

Amplitude: 3
Period: $\pi$

#### Explanation:

graph{y=3sin(2x) [-5.46, 14.54, -4.28, 5.72]}

Considering the $x$-axis to be the time as the wave progresses and the $y$-axis the displacement, the period can be seen to be $\pi$ as that is the time taken for one complete oscillation

(Proof)

$y = 0$

$0 = 3 \sin \left(2 t\right)$

$0 = \sin \left(2 t\right)$

$2 t = {\sin}^{-} 1 \left(0\right)$

$2 t = 0 , \pi , 2 \pi , 3 \pi , 4 \pi \ldots$

$t = 0 , \frac{1}{2} \pi , \pi , \frac{3}{2} \pi , 2 \pi \ldots$

(every second time the wave crosses the $x$-axis is a complete oscillation)

$t = 0 , \pi , 2 \pi \ldots$

The constant difference is $\pi$
$\therefore$ the period is $\pi$

The amplitude is the maximum distance from the rest point of a wave - this is the same as the maximum value of the graph.

$y = \sin x$ has a range of $- 1 \le y \le 1$
(the maximum value possible is 1)

This makes the maximum value of $3 \sin \left(2 t\right) = 3 \left(1\right) = 3$
$\therefore$ the amplitude is 3