How do you find the amplitude and period of #y= 4/3 sin (2/3)x#?

1 Answer
Jul 12, 2015

This question has no sense to me, the x should be INSIDE the sinusfor being a periodic function. If the x is really out the parentesis then is a tricky question. It has no period nor amplitude..

Explanation:

If you really wrote it correctly then #y=4/3sin(2/3)x=0.824x# ot's a linnear equation and it has no period nor amplitude, it's a linnear function.

If your equation is #y=4/3sin(2/3x)# then you need to know that the sin oscillates between -1 and 1 so if it's multipliyed by 4/3 your equaion is oscillating between 4/3 and -4/3 so you have you amplitude, it's 4/3.

for the period it is #3pi#, for x=0 the equation's value is 4/3, for #y=4/3sin(2pi)# you again have y=4/3 betwwen this two point there's no other point with x=4/3 so between these two has passed a period.

The difference between the x's will be the period, then isolating x form #2pi=2/3x# you get #x=3pi# and #3pi-0=3pi# then #T=3pi#