anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
so one period is 1.5 right? the other is 1.5 right?
anonymous
  • anonymous
yup
amistre64
  • amistre64
2pi ---- = 1.5; w = 2pi/1.5 right? w

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anonymous
  • anonymous
yup, w=4.189
amistre64
  • amistre64
without knowing anything more about the wave, or the graph there is no wayto determine the amplitude
amistre64
  • amistre64
since y=0 when the period happens we can assume its not lifted
amistre64
  • amistre64
since it is symetric about the orgin it aint shifted either
anonymous
  • 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
  • amistre64
giving all the information to a question is usually a good way to get an informed answer ;)
anonymous
  • anonymous
hahha ok one second and I will type the points out.
anonymous
  • anonymous
A=(Max - min)/2
anonymous
  • 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
  • anonymous
(0,0) use sine and no phase shift
anonymous
  • anonymous
\[.75sin(4.189x)\]?
anonymous
  • anonymous
how did you figure "a" as .75?
anonymous
  • anonymous
oh because i was wrong. sorry
anonymous
  • anonymous
range from -.75 to 2.25 yes?
anonymous
  • anonymous
yup
anonymous
  • anonymous
i have to write hold on
anonymous
  • anonymous
ok
anonymous
  • anonymous
range has length 3 yes?
anonymous
  • anonymous
yes
anonymous
  • anonymous
so you need 1.5 out front
anonymous
  • anonymous
so it will have range of length 3
anonymous
  • anonymous
then it goes from -.75 to 2.25 so at the end you will need to add .75
anonymous
  • 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
  • anonymous
so far we know it looks like \[y= 1.5 sin(___) +.75\]
anonymous
  • anonymous
now we have to figure out what goes inside.
anonymous
  • anonymous
b=4.189
anonymous
  • anonymous
i meant \[y=1.5sin ( ?)+.75\]
anonymous
  • anonymous
period = 2pi/b and period we found was 1.5 right? so solve for b and its 4.189
anonymous
  • anonymous
that takes care of period yes?
anonymous
  • anonymous
wait let me check hold on
anonymous
  • anonymous
ok
anonymous
  • 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
  • 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
  • anonymous
does that link work if not i can find another
anonymous
  • anonymous
http://www.ltcconline.net/greenl/courses/204/appsHigherOrder/forcedVibrations.htm
anonymous
  • anonymous
that links better scroll down to the graph, see it?
anonymous
  • anonymous
oh man i was way way off base. the last graph?
anonymous
  • anonymous
where it says y=tsin(10t)? that graph
anonymous
  • anonymous
not the graph at the very bottom, where it says y=tsin(10t)
anonymous
  • anonymous
ah i see it
anonymous
  • 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
  • anonymous
so amplitude changes
anonymous
  • anonymous
yes and period too
anonymous
  • anonymous
oh lord i have no friggin idea how to model this. let me think
anonymous
  • 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
  • anonymous
well there is no exponential part of the picture you just sent. that is the last graph
anonymous
  • anonymous
if you connected just the maximum values it is..
anonymous
  • anonymous
are you looking at the right graph its the graph right before "Forced Vibrations with Damping"
anonymous
  • anonymous
i have to confess that i really have no idea how to do this
anonymous
  • anonymous
oh ok..

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