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Easy Simple Harmonic Motion Problem

  1. Apr 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.5 m. Each particle has a period of 1.5 s, but they differ in phase by π/5 rad. (a) How far apart are they 0.46 s after the lagging particle leaves one end of the path? (b) Are they then moving in the same direction, toward each other, or away from each other?

    2. Relevant equations
    x=a*cos(wt+ø)
    T=2π/w
    3. The attempt at a solution
    the amplitude is given by a=1.5m
    and T=1.5s
    then we have w=2π/T=4.18879..
    the two particles are identical except they differ in phase by π/5
    my attempt in finding the distance between them after t=0.46s is
    d=1.5cos(wt)-1.5cos(wt+π/5)=0.28647
    which does not match the correct solution
    can someone point out where I went wrong
     
  2. jcsd
  3. Apr 30, 2016 #2

    ehild

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    "How far apart are they 0.46 s after the lagging particle leaves one end of the path?"
    At t=0, the lagging particle is just leaving one end of the path, say the left end, where x1=0, and turning to move toward the other end at x2=1.5 m. What is the function x(t) then?
     
  4. Apr 30, 2016 #3
    Then x(t)=1.5 this means cos(¢)=1
     
  5. Apr 30, 2016 #4
    pls ignore my previous post,
    x(0)=0=1.5cos(0+pi/2)
     
  6. Apr 30, 2016 #5

    ehild

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    It is "simple harmonic motion along a common straight-line segment of length 1.5 m." That can mean from 0 to 1.5, for example, or from -0.75 to 0.75, as an other example. What is the amplitude?
     
  7. Apr 30, 2016 #6
    I see... 0.75 it is
     
  8. Apr 30, 2016 #7

    ehild

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    Yes. And what is the x(t) function, so as it is at the left end of the interval at t=0 and the velocity just turning to positive?
     
  9. Apr 30, 2016 #8
    So for when its at the left end where x=0 at t=0,
    x(0)=0.75cos(π/2) ===> ø=π/2
    x(t)=0.75cos(wt+π/2)=0.75sin(wt)
    and the other particle has
    x'(t)=0.75cos(wt+π/5+π/2)=sin(wt+π/5)
     
  10. Apr 30, 2016 #9

    ehild

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    x(t) = 0.75sin(wt) varies between -0.75 and 0.75, and at t=0 it is at x=0. It is not one end of the line segment.
    Decide where are the end of the light segment first.
     
  11. Apr 30, 2016 #10
    okay so at one end x=-0.75=0.75sin(wt), wt=3π/2
     
  12. Apr 30, 2016 #11

    ehild

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    Well, where are both particles 0.46 s later? How far are they?
     
  13. Apr 30, 2016 #12
    X=sin(0.46w)=0.937282
     
  14. Apr 30, 2016 #13

    ehild

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    The question was "How far apart are they 0.46 s after the lagging particle leaves one end of the path?". The lagging particle leaves the left end when wt = 3π/2. Where is it 0.46 s later?
     
  15. May 1, 2016 #14
    when it leaves the left end wt=3π/2 ==> t=3π/2w
    0.46s later t'=3π/2w + 0.46
    the position is x=sin(wt')=sin(3π/2+0.46w)
     
  16. May 1, 2016 #15

    ehild

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    Yes, and what is the position of the other particle at the same time? What is the distance between them?
     
  17. May 1, 2016 #16
    the other particle has position x=sin(3π/2+0.46w+π/5) since they differ in phase by π/5
    the distance would be the difference between the two x values
     
  18. May 1, 2016 #17

    ehild

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    Yes, but you forgot the amplitude from both x-s.
    Calculate the distance.
     
  19. May 1, 2016 #18
    oh yes it should be
    d=0.75sin(3π/2+0.46w+π/5)-0.75sin(3π/2+0.46w)=0.363261895
     
  20. May 1, 2016 #19

    ehild

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    OK, it is correct.
    Now question b) :)
     
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