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Amplitude in pendulum motion

  1. Nov 18, 2007 #1
    Amplitude in pendulum motion....

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    w = [tex]\frac{g}{L}[/tex]

    T = 2[tex]\pi[/tex][tex]\sqrt{\frac{L}{g}}[/tex]

    I[tex]\left(\frac{d^{2}\theta}{dt^{2}}\right)[/tex] = -mgL[tex]\theta[/tex]

    x(t) = Acos(wt) ??

    3. The attempt at a solution

    I'm stuck on this problem. I haven't worked with harmonic motion enough yet to grasp it. Any help would be greatly appreciated.

    I need a hand on this.
  2. jcsd
  3. Nov 18, 2007 #2


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    At the lowermost position of the mass of the pendulum the tension on the string is equal to the centripetal reaction. i.e. T= mv^2/r. You can find the velocity by using the conservation of energy. i.e. K.E = P.E.
  4. Nov 18, 2007 #3

    Shooting Star

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    So, where are you putting the weight of the mass at the lowermost position?
  5. Nov 18, 2007 #4


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    In the pendulum the mass is swinging. During the swing when the mass crosses the lowermost point, the tension in string is maximum and it is equal to mv^2/r + mg According to the conservation of energy 1/2*m*v^2 = mgh, where h is the height from mthe lower point from which the mass of the pendulum is released. If you know you can find the amplitude.
    Last edited: Nov 18, 2007
  6. Nov 18, 2007 #5

    Doc Al

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    Staff: Mentor

    Incorrect. There are two forces acting on the mass at that point; tension is one of them. (This was Shooting star's point.) The net force equals mv^2/r.

  7. Nov 18, 2007 #6


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    Right. Tension = mv^2/r + mg.
  8. Nov 18, 2007 #7
    wow, so it turns out that we don't need to use any concepts from this particular chapter. I was actually onto this the way you guys described it, but then i said to myself that we must have to use some formula or something from this chapter. I guess not.

    T = [tex]\frac{mv^{2}}{r}[/tex] + mg --->

    v = [tex]\sqrt{r\left(\frac{T}{m} - g\right)}[/tex]

    v = 1 m/s --->

    E[tex]_{i}[/tex] = E[tex]_{f}[/tex]

    mgh = [tex]\frac{1}{2}[/tex]mv[tex]^{2}[/tex] --->

    h = [tex]\frac{v^{2}}{2g}[/tex] --->

    h = 0.05m --->

    then from my free body diagram

    0.45 - 0.05 = 0.40m

    cos[tex]^{-1}[/tex][tex]\left(\frac{0.40}{0.45}\right)[/tex] = 27.27[tex]^{o}[/tex]

    The textbook answer is 27.16[tex]^{0}[/tex]. I always get answers that are a little off from the text. Calculators?
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