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Vertical Vibrations and Lissajous curve

  1. Nov 26, 2012 #1
    1. The problem statement, all variables and given/known data

    a)x=cos2ωt, y=sin2ωt
    b)x=cos2ωt, y=cos(2ωt-∏/4)
    c)x=cos2ωt, y=cosωt

    draw the graphs of Simple Harmonic Motions.


    2. Relevant equations

    parametric and complex harmonic motion equation is needed.


    3. The attempt at a solution

    No attempt to solution.I can't do anything,because I can't understand what does all it mean.
    It's not my homework.I'm just preparing to my exam.So can anyone tell me step by step how can I draw these graphs and what I have to know about these graphs? They say that we can find the frequency of motion with this but how?? it's my question..

    Sorry for my bad English,
     
    Last edited: Nov 26, 2012
  2. jcsd
  3. Nov 27, 2012 #2

    haruspex

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    To sketch a curve given in parametric form:
    - Find the max and min values of x and y, and the values of t for which these occur.
    - For each of these, whether it be an extremum of x or y, plot its (x,y) position.
    - Look for intersection points, i.e. (x,y) points that arise from different values of t (other than those that arise from the periodicity). Plot those. Might also help to figure out the gradients at these (two or more gradients at each intersection point).
    - Generally you'd also look for asymptotes, but clearly x and y are bounded here.
    With all that laid out, you should be able to connect the dots.
     
  4. Nov 27, 2012 #3
    Can I use this method to find x and y?

     
    Last edited by a moderator: Sep 25, 2014
  5. Nov 27, 2012 #4

    haruspex

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    That really isn't a good way to go about it. In curve sketching, you should be aiming to extract the interesting aspects: maxima, minima, inflexions, asymptotes, self-intercepts, whether it goes through the origin, and maybe other axis intercepts. You can't expect to achieve that just by plotting arbitrary points.
    How would you find a local maximum y given a parametric form?
     
    Last edited by a moderator: Sep 25, 2014
  6. Nov 27, 2012 #5
    You're right.But I don't have a enough time so I'm looking for shortway to do this.
    Because I can't remember how we find the minimum and maximum points on parametric forms?

    There's another way;
    http://tutorial.math.lamar.edu/Classes/CalcII/ParametricEqn.aspx

    Example 5.

    Or can you help me to graph the first function.It will be good for me to get the idea.

    Thanks for help!
     
  7. Nov 27, 2012 #6

    haruspex

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    You just differentiate x and y wrt t. The ratio gives dy/dx (or dx/dy by inverting the ratio). So a local max or min of y is where dy/dt = 0, and a local max or min of x is where dx/dt = 0. Figuring out whether max or min is a little trickier, but that should become apparent as you build up the sketch. Plotting one or two specific points can remove any ambiguity.
    Elimination of the parameter can only be done in special cases. The first in the OP is an example, so is the second though a little harder. You do have to be careful, though. Consider x = cos t, y = cos t. Elimination produces y = x, but values of each outside [-1,1] are not possible.
    OK, but in the interests of being able to use the method generally I won't do it by elimination.
    dy/dt = 2 cos(2ωt) = 0 gives 2ωt = (n+1/2)π, for which (x,y) = (0,±1).
    dx/dt = -2 sin(2ωt) = 0 gives 2ωt = nπ, for which (x,y) = (±1,0).
    In connecting up these extrema, there is still an ambiguity. E.g. in going from (-1,0) to (0,1), does it curve up and to the right or does it go up and to the left off to some asymptote, then reappear from (-∞,+∞), coming down and to the right to reach (0,1)? To answer that, you can either plot some intermediate point or simply observe that cos and sin are each limited to being in the range [-1,1].
     
  8. Nov 27, 2012 #7
    Ok.You're right,I have to understand the general mode.But I can't find any documentation how to do all these on any trigonometric function.All of them are polynoms.

    Is it the graph of function?

    25sbymd.jpg
     
  9. Nov 27, 2012 #8

    haruspex

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    Have you tried applying my method to (b) and (c)?
    Yes. You can see this very easily by eliminating the parameter and getting x2+y2=1.
     
  10. Nov 27, 2012 #9
  11. Nov 27, 2012 #10

    haruspex

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    I've already explained how to find the x and y extrema by differentiating wrt the parameter. Try my method for (b). If you can't complete it, post whatever you get.
    If you want to find the intercepts on an axis, x=0 say, plug x=0 into the parametric equation for x and determine the possible values of the parameter. Then plug those into the equation for y and you will have the intercepts.
     
  12. Nov 27, 2012 #11
    Ok.I'm trying to do for c)

    x=cos2ωt y=cosωt

    dx/dt = -2sin(2ωt) = 0 (±1,0)
    dy/dt = -sin(ωt) = 0 (0,0) We are looking where ωt is zero?

    Now shape is like parabola on Y coordinates?
     
  13. Nov 27, 2012 #12
    b)

    points are;

    (±1,0) and (0,0.4) ?
     
  14. Nov 27, 2012 #13

    haruspex

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    There are more solutions. sin(2ωt) = 0 iff 2ωt=nπ for some integer n.
    x = cos (nπ) = ±1
    y = cos (nπ/2) = ?
    Again, there are more solutions. Need to consider all.
    I wouldn't think so. Can |x| or |y| exceed 1?
     
  15. Dec 1, 2012 #14
    :) I come back to here.I'm sorry,but I have a problem with language.I can't understand everythink in English so sometimes I have to read it repeatedly.

    Here is what I have tried to b) again :)

    dy/dt = -2sin(2ωt-∏/2)---> 2ωt = ((n+1/2)∏-∏/4) n=0 -> ∏/4 = 0.707, n=1 ->-∏/4 = -0.707
    dx/dt = -2sin(2wt) ---> 2wt= (n+1/2)∏ --> n =0,1,2,3 (-1,1)

    dy/dt = (0,±0.707)
    dx/dt = (±1,0)

    So the graph is look like;


    Edit : there are a mistakes.
     

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    Last edited: Dec 1, 2012
  16. Dec 1, 2012 #15

    (-1,+1,0)
     
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