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Simple Harmonic/Spring Question

  1. Feb 10, 2005 #1
    http://www.quantumninja.com/hw/random/34.JPG

    I was hoping someone could help me with those two problems

    I am not sure how to start 3. Here is what I have done with 4 so far...

    [tex] \sum F= -kx-mg [/tex]

    [tex] k=245 [/tex]

    [tex] \sum F= -kx-mg [/tex]

    [tex] \sum F= 34.3 [/tex]

    F=ma
    [tex] F=m x(double prime) [/tex] where [tex] x= Acos(\omega t + \theta) [/tex]

    after that I am lost

    Please help
    thx
     
  2. jcsd
  3. Feb 10, 2005 #2

    dextercioby

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    HINT for 4.U don't need the differential equation.Nor the acceleration of gravity.

    Daniel.
     
  4. Feb 11, 2005 #3
    Are you doing this for a course in physics, or differential equations?

    If it's for an elementary physics course, as dextercioby said, solving that differential equation is a lot more work than you need to do.

    The period of a simple harmonic oscillator is given by

    [tex]T = \frac{2 \pi}{\omega}[/tex]

    and [tex]\omega = \sqrt{ \frac{k}{m}}[/tex]

    =======================================================

    As for #3, put your calculator away & visualize the motion of the particles. Slide your fingers around if necessary. Concentrate on the meanings of frequency, amplitude, displacement, and phase difference.

    Think, what do they mean by "each time their displacement is half their amplitude". When does that happen?
     
    Last edited: Feb 11, 2005
  5. Feb 11, 2005 #4
    ok for number 4 according to what I believe you meant

    is it just

    [tex] \omega = \sqrt{490} [/tex]

    and therefore

    [tex] T= \frac{2\pi}{\sqrt{490}} [/tex]
     
  6. Feb 11, 2005 #5
    For number 3 I have been looking at the equation

    [tex]x=X_{m}cos(\omega t + \phi)[/tex]

    Am I going about this the right way. I know there should be a simple set here, but I am missing it
     
  7. Feb 11, 2005 #6
    Stop looking at equations. What does this mean:
    "They pass each other going in opposite directions each time their displacement is half their amplitude."

    It says the frequencies and amplitudes are equal, right? Let's say you have two particles A and B and the period of each one is 4 seconds (just making up a number) and the amplitude is just the length of this:

    [tex]================================[/tex]​

    Redraw that a few times & try marking on each one the positions of the two particles at each time for t=0, t=1, t=2, t=3, t=4 (seconds)
     
  8. Feb 11, 2005 #7
    [tex]\frac{2\pi}{3}[/tex]
     
  9. Feb 11, 2005 #8
    I am getting [tex]\pi[/tex]
     
  10. Feb 11, 2005 #9
    Tom, it's hard to help you if I don't know how you're thinking about it.

    Suppose the particles are oscillating with a frequency of .25 cycle/s

    Suppose at time 0 two particles A and B are at opposite ends of the line:

    T=0...A========================B


    Where will they be at these times:


    T=1s....========================

    T=2s....========================

    T=3s....========================

    T=4s....========================
     
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