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Showing SHM

  1. Feb 28, 2014 #1
    1. The problem statement, all variables and given/known data

    A cube of mass m is connected to two rubber bands of length L, each under tension T. The cube is displaced by a small distance y perpendicular to the length of the rubber bands. Assume the tension doesn't change. Show that the system exhibits SHM, and find its angular freqency ω.

    3. The attempt at a solution

    So basically from a FBD of cube, I have vertical forces: -2Tsinθ - mg = m[itex]\frac{d^2y}{dt^2}[/itex] and the horizontal components of tension from each band cancels. Now since the cube is displaced by a small distance y, I assume we can approximate sinθ ≈ θ. But then I'm not sure what to do?
    I tried using sinθ = [itex]\frac{y}{(y^2+L^2)^{1/2}}[/itex], but then I get a complicated expression.
    I know I need to obtain a -constant*y on the LHS. Any suggestions.
     
  2. jcsd
  3. Feb 28, 2014 #2
    I think i got it, your assuming sinθ ≈ y/L, for small y.
    One question though, do we have to assume gravity is negligible to get a sensible answer?
     
    Last edited: Feb 28, 2014
  4. Feb 28, 2014 #3

    vela

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    No, you don't. The differential equation becomes
    $$y'' + \frac{2T}{mL}y + g=0.$$ Now consider a change of variables to ##u = y+\frac{mg}{2T}L##. What's the differential equation in terms of ##u##? What does ##\frac{mg}{2T}L## physically represent?
     
  5. Feb 28, 2014 #4
    We haven't studied differential equations in much depth (since its a introductory physics course), so I didn't really catch the change of variables part. If you could explain further, I'd appreciate it.
     
  6. Feb 28, 2014 #5
    I would just throw gravity out. The problem doesn't specifically mention it, so it might as well be on a frictionless tabletop or so.
     
  7. Mar 1, 2014 #6

    vela

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    I'm saying rewrite the equation in terms of u instead of y. If you differentiate u twice, you get u''=y'', right? Just substitute in for y and y''.

    It would be kind of hard to oscillate vertically on a flat tabletop.
     
  8. Mar 1, 2014 #7
    When I read the problem, I also assumed (as jackarms did) that the oscillation was taking place in the horizontal, not the vertical. There is nothing in the problem statement that mentions the vertical. I pictured a horizontal frictionless table.

    Chet
     
  9. Mar 1, 2014 #8
    this is what I came out with...I also went on to show the frequency as well...
     

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  10. Mar 1, 2014 #9
    This is definitely not what my answer would have been. I would have had T as a parameter in the frequency. I would have had:
    [tex]f=\frac{1}{2π}\sqrt{\frac{2T}{mL}}[/tex]
    This is based on Vela's equation in post #3, with g removed.

    Chet
     
  11. Mar 1, 2014 #10

    vela

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    That's true. I inferred it from the use of ##y##, but that's not really justified.
     
  12. Mar 2, 2014 #11
    The angular frequency is actually represented by omega...f is the frequency of oscillation
     
  13. Mar 2, 2014 #12
    Yes. ω=2πf

    Chet
     
  14. Mar 2, 2014 #13
    Yip so my proof of the angular frequency stops at the third to last line
     
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