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so one period is 1.5 right?
the other is 1.5 right?

yup

2pi
---- = 1.5; w = 2pi/1.5 right?
w

yup, w=4.189

without knowing anything more about the wave, or the graph there is no wayto determine the amplitude

since y=0 when the period happens we can assume its not lifted

since it is symetric about the orgin it aint shifted either

giving all the information to a question is usually a good way to get an informed answer ;)

hahha ok one second and I will type the points out.

A=(Max - min)/2

(0,0) use sine and no phase shift

\[.75sin(4.189x)\]?

how did you figure "a" as .75?

oh because i was wrong. sorry

range from -.75 to 2.25 yes?

yup

i have to write hold on

ok

range has length 3 yes?

yes

so you need 1.5 out front

so it will have range of length 3

then it goes from -.75 to 2.25 so at the end you will need to add .75

so far we know it looks like \[y= 1.5 sin(___) +.75\]

now we have to figure out what goes inside.

b=4.189

i meant \[y=1.5sin ( ?)+.75\]

period = 2pi/b and period we found was 1.5 right? so solve for b and its 4.189

that takes care of period yes?

wait let me check hold on

ok

does that link work if not i can find another

http://www.ltcconline.net/greenl/courses/204/appsHigherOrder/forcedVibrations.htm

that links better scroll down to the graph, see it?

oh man i was way way off base. the last graph?

where it says y=tsin(10t)? that graph

not the graph at the very bottom, where it says y=tsin(10t)

ah i see it

so amplitude changes

yes and period too

oh lord i have no friggin idea how to model this. let me think

well there is no exponential part of the picture you just sent. that is the last graph

if you connected just the maximum values it is..

are you looking at the right graph its the graph right before "Forced Vibrations with Damping"

i have to confess that i really have no idea how to do this

oh ok..