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anonymous

  • 5 years ago

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.

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  1. amistre64
    • 5 years ago
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    so one period is 1.5 right? the other is 1.5 right?

  2. anonymous
    • 5 years ago
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    yup

  3. amistre64
    • 5 years ago
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    2pi ---- = 1.5; w = 2pi/1.5 right? w

  4. anonymous
    • 5 years ago
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    yup, w=4.189

  5. amistre64
    • 5 years ago
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    without knowing anything more about the wave, or the graph there is no wayto determine the amplitude

  6. amistre64
    • 5 years ago
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    since y=0 when the period happens we can assume its not lifted

  7. amistre64
    • 5 years ago
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    since it is symetric about the orgin it aint shifted either

  8. anonymous
    • 5 years ago
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    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?

  9. amistre64
    • 5 years ago
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    giving all the information to a question is usually a good way to get an informed answer ;)

  10. anonymous
    • 5 years ago
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    hahha ok one second and I will type the points out.

  11. anonymous
    • 5 years ago
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    A=(Max - min)/2

  12. anonymous
    • 5 years ago
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    (-.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)

  13. anonymous
    • 5 years ago
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    (0,0) use sine and no phase shift

  14. anonymous
    • 5 years ago
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    \[.75sin(4.189x)\]?

  15. anonymous
    • 5 years ago
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    how did you figure "a" as .75?

  16. anonymous
    • 5 years ago
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    oh because i was wrong. sorry

  17. anonymous
    • 5 years ago
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    range from -.75 to 2.25 yes?

  18. anonymous
    • 5 years ago
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    yup

  19. anonymous
    • 5 years ago
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    i have to write hold on

  20. anonymous
    • 5 years ago
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    ok

  21. anonymous
    • 5 years ago
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    range has length 3 yes?

  22. anonymous
    • 5 years ago
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    yes

  23. anonymous
    • 5 years ago
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    so you need 1.5 out front

  24. anonymous
    • 5 years ago
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    so it will have range of length 3

  25. anonymous
    • 5 years ago
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    then it goes from -.75 to 2.25 so at the end you will need to add .75

  26. anonymous
    • 5 years ago
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    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)

  27. anonymous
    • 5 years ago
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    so far we know it looks like \[y= 1.5 sin(___) +.75\]

  28. anonymous
    • 5 years ago
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    now we have to figure out what goes inside.

  29. anonymous
    • 5 years ago
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    b=4.189

  30. anonymous
    • 5 years ago
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    i meant \[y=1.5sin ( ?)+.75\]

  31. anonymous
    • 5 years ago
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    period = 2pi/b and period we found was 1.5 right? so solve for b and its 4.189

  32. anonymous
    • 5 years ago
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    that takes care of period yes?

  33. anonymous
    • 5 years ago
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    wait let me check hold on

  34. anonymous
    • 5 years ago
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    ok

  35. anonymous
    • 5 years ago
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    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?

  36. anonymous
    • 5 years ago
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    does that link work if not i can find another

  37. anonymous
    • 5 years ago
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    http://www.ltcconline.net/greenl/courses/204/appsHigherOrder/forcedVibrations.htm

  38. anonymous
    • 5 years ago
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    that links better scroll down to the graph, see it?

  39. anonymous
    • 5 years ago
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    oh man i was way way off base. the last graph?

  40. anonymous
    • 5 years ago
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    where it says y=tsin(10t)? that graph

  41. anonymous
    • 5 years ago
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    not the graph at the very bottom, where it says y=tsin(10t)

  42. anonymous
    • 5 years ago
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    ah i see it

  43. anonymous
    • 5 years ago
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    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)

  44. anonymous
    • 5 years ago
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    so amplitude changes

  45. anonymous
    • 5 years ago
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    yes and period too

  46. anonymous
    • 5 years ago
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    oh lord i have no friggin idea how to model this. let me think

  47. anonymous
    • 5 years ago
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    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

  48. anonymous
    • 5 years ago
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    well there is no exponential part of the picture you just sent. that is the last graph

  49. anonymous
    • 5 years ago
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    if you connected just the maximum values it is..

  50. anonymous
    • 5 years ago
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    are you looking at the right graph its the graph right before "Forced Vibrations with Damping"

  51. anonymous
    • 5 years ago
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    i have to confess that i really have no idea how to do this

  52. anonymous
    • 5 years ago
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    oh ok..

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