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Question regarding Simple Harmonic Motion

  1. Dec 12, 2012 #1
    1. The problem statement, all variables and given/known data

    The two linear simple harmonic motions of equal amplitudes , and angular frequencies ω and 2ω are imposed on a particle along the axes of X and Y respectively. If the initial phase difference between them is π/2 , then find the resultant path followed by the particle.


    2. Relevant equations

    http://en.wikipedia.org/wiki/Simple_harmonic_motion

    3. The attempt at a solution

    I tried solving question using x= a sinωt
    And other is y=a sin(2ωt + π/2)
    Solving them , I get the wrong answer in terms of x and y....

    Please help !!

    Thanks in advance...
    :smile:
     
  2. jcsd
  3. Dec 12, 2012 #2

    gneill

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    Staff: Mentor

    Can you show how you went about "solving them"? We need to see your attempt if we're to know how to help.
     
  4. Dec 12, 2012 #3

    ehild

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    Gold Member

    Hi sankalpmittal,

    Show please what you have tried?
    x=sin(ωt) and y=sin(2ωt+π/2) =cos(2ωt) is the parametric representation of a Lissajous curve. See http://en.wikipedia.org/wiki/Lissajous_curve

    Try to eliminate the ωt term.

    ehild
     
  5. Dec 13, 2012 #4
    Firstly I solved this question using :
    a : Amplitude
    x=a cosωt
    y=a sin(2ωt+π/2)
    y=a cos (2ωt)
    y= a( cos2ωt-sin2ωt)
    y= a(2cos2ωt-1)
    y= a{(2x2/a2)-1)

    I noticed that this does not match the answer given in my textbook...

    Then I made a second attempt :
    x=a sinωt
    y=a sin (2ωt+π/2)
    y=a cos(2ωt)
    y= a(1-2sin2ωt)
    y= a{1-(2x2/a2)}

    Again this does also not match with the answer in my textbook....

    Then I made the third attempt :

    x= a cosωt
    y= a cos(2ωt+π/2)
    y= -asin(2ωt)
    y= -2a sin(ωt)cos(ωt)
    y=-2a √(1-sin2ωt) (x/a)
    y=-2a √{1-(x2/a2)} (x/a)
    On squaring both sides and simplifying , I got :

    y= 4x2{1-(x2/a2)}

    This answer matched with the answer given in my textbook....

    I just wanted to know , where I went wrong in my first and second attempt....
     
  6. Dec 13, 2012 #5

    ehild

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    They are ins phase, instead of being shifted by pi/2.

    .

    That is correct, if x and y are as you assumed.

    But this answer is wrong as y is not squared. It was correct before squaring.

    If you add two perpendicular vibrations you can have different y(x) curves, depending on the initial phases.


    ehild
     
  7. Dec 15, 2012 #6
    OK , so there was a typo. The answer I got in my third attempt was y2=4x2{1-(x2/a2)} , which matched with the answer given in my textbook. But unfortunately the answer which I got in my first and second attempt , did not match with that given in my textbook.

    According to you , both the equations , in my second and third attempt are correct , yet they are different and only latter match with the answer in my textbook. How ?

    Also what do you mean by "in phase and not shifted by pi/2" in my first attempt ? Can you explain a bit comprehensively ?
     
  8. Dec 15, 2012 #7

    ehild

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    Your first attempt was
    x=a cosωt
    y=a sin(2ωt+π/2) which is equivalent to
    y=a cos (2ωt)I do not see any initial phase difference between x and y.
    Your second attempt is true and so is the third one. Phase difference between x and y is not a clear concept. You get x,y curves of different form if x=sin(ωt), y=cos(2ωt) (y=1-2x2) and when x=cos(ωt) and y=sin(2ωt) (y2=4x2(1-x2))

    ehild
     

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  9. Dec 15, 2012 #8

    SammyS

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    A better link to consider might be: http://en.wikipedia.org/wiki/Lisajous .

    Also see: http://en.wikipedia.org/wiki/Lemniscate_of_Gerono .
     
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