I need help finding the equation of a sine function, I know (-.75,0) to (.75,0) is one period and (.75,0) to (2.25,0) is another so based on that I was able to solve for "b" by applying the equation Period = 2pi/b so in that case 1.5=2pi/b, and therefore b=4.189, im stuck finding "a" and if there was any horizontal shift left or right, or any vertical shift up or down.

- anonymous

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- amistre64

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

- anonymous

yup

- amistre64

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

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## More answers

- anonymous

yup, w=4.189

- amistre64

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

- amistre64

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

- amistre64

since it is symetric about the orgin it aint shifted either

- anonymous

well, can I give you the rest of the points? it's a sine function essentially but with changing amplitude and period so my teacher suggested first getting the equation for the maximum values and then for a period piece and adding the two together does that make sense?

- amistre64

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

- anonymous

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

- anonymous

A=(Max - min)/2

- anonymous

(-.75,0,) ...(-.5, -.7906) ....(0,0) ..... (.25, .90461),.... (.5, .94868) ....(.75, 0) ...(1, -1.039),... (1.25, -1.088) ...(1.5,0).... (1.75, 1.1915) .....(2,1.2471) and (2.25, 0)

- anonymous

(0,0) use sine and no phase shift

- anonymous

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

- anonymous

how did you figure "a" as .75?

- anonymous

oh because i was wrong. sorry

- anonymous

range from -.75 to 2.25 yes?

- anonymous

yup

- anonymous

i have to write hold on

- anonymous

ok

- anonymous

range has length 3 yes?

- anonymous

yes

- anonymous

so you need 1.5 out front

- anonymous

so it will have range of length 3

- anonymous

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

- anonymous

this will take you from - .75 when the function is -1 (or rather -1.5) up to 2.25 when the function is 1 (or 1.5)

- anonymous

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

- anonymous

now we have to figure out what goes inside.

- anonymous

b=4.189

- anonymous

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

- anonymous

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

- anonymous

that takes care of period yes?

- anonymous

wait let me check hold on

- anonymous

ok

- anonymous

oh no i am all messed up. i had the range at 3 but that is wrong wrong wrong. the range goes from smallest y value to biggest. what is the smallest y value?

- anonymous

ok here maybe I can explain it better because I am trying to break this up as my teacher suggested, the full graph of what I need the equation for looks like the green graph on this link: http://www.google.com/imgres?imgurl=http://fan.lib.ru/img/z/zharow_a/evolution_continues/majatnik_kachajasx_bez_trenija.jpg&imgrefurl=http://fan.lib.ru/z/zharow_a/evolution_continues.shtml&usg=__uANt4T4noENBP-ROlSIgJEg9idI=&h=253&w=787&sz=51&hl=en&start=0&zoom=1&tbnid=L3R1ovN3yt9ifM:&tbnh=67&tbnw=208&ei=2RDkTc2gDsb00gHgms2cDg&prev=/search%3Fq%3Dgrowing%2Bamplitude%26um%3D1%26hl%3Den%26biw%3D1080%26bih%3D535%26tbm%3Disch&um=1&itbs=1&iact=hc&vpx=74&vpy=231&dur=243&hovh=127&hovw=396&tx=210&ty=90&page=1&ndsp=15&ved=1t:429,r:5,s:0&biw=1080&bih=535

- anonymous

does that link work if not i can find another

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ah i see it

- anonymous

yeah so thats the shape of what I am working with and I given the following 12 points: (-.75,0,) ...(-.5, -.7906) ....(0,0) ..... (.25, .90461),.... (.5, .94868) ....(.75, 0) ...(1, -1.039),... (1.25, -1.088) ...(1.5,0).... (1.75, 1.1915) .....(2,1.2471) and (2.25, 0)

- anonymous

so amplitude changes

- anonymous

yes and period too

- anonymous

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

- anonymous

well what my teacher recommended is finding the exponential equation of the maximum values and then finding the equation for a period piece like we were just trying to do and then adding the two functions together if that makes sense if not we can try any other way yoou can come up withh

- anonymous

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

- anonymous

if you connected just the maximum values it is..

- anonymous

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

- anonymous

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

- anonymous

oh ok..

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