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Approximation of SHM

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data
    The restoring force of a pendulum is [itex]F_\theta=-mg\sin\theta[/itex]
    and is approximated to [itex]F_\theta=-mg\theta[/itex].
    The period is [itex]T=2\pi\sqrt{\frac{L}{g}}[/itex], but can be expressed as the infinite series:
    [tex]T=2\pi\sqrt{\frac{L}{g}}\left( 1+\frac{1^2}{2^2}\sin^2\frac{\theta}{2}+\frac{1^2}{2^2}\frac{3^2}{4^2}\sin^4\frac{\theta}{2}+...\right)[/tex]
    What is this approximation and of what? i don't think it's a Maclaurin series.

    2. Relevant equations
    Maclaurin series of sin(x):
    [tex]\sin(x)\cong 1-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...[/tex]
     
  2. jcsd
  3. Apr 29, 2014 #2

    Doc Al

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    Staff: Mentor

    When solving for the period of a large amplitude pendulum you get a nasty elliptical integral. You can express that integral in terms of a series, the first term being the familiar period for small amplitudes.

    Try this for more details: LARGE-ANGLE MOTION OF A SIMPLE PENDULUM
     
  4. May 5, 2014 #3
    Change of variables

    In the link you gave me:
    http://api.viglink.com/api/click?fo...M&txt=LARGE-ANGLE MOTION OF A SIMPLE PENDULUM
    there is equation 7:
    [tex]\sin(\varphi)=\frac{\sin(\vartheta/2)}{\sin(\alpha/2)}[/tex]
    It says [itex]\alpha[/itex] changes from 0 to 2[itex]\pi[/itex] for a full oscillation.
    First, when [itex]\alpha=0[/itex] then the denominator=0
    Secondly, [itex]\alpha[/itex] changes from [itex]-\vartheta[/itex] to [itex]+\vartheta[/itex]
    so [itex]\sin(\varphi)[/itex] changes from [itex]-\varphi[/itex] [itex]+\varphi[/itex]
     
  5. May 5, 2014 #4

    Doc Al

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    Staff: Mentor

    I believe it is [itex]\varphi[/itex] that ranges from 0 to 2[itex]\pi[/itex] for a full oscillation, not [itex]\alpha[/itex]. [itex]\alpha[/itex] is the initial angle of the pendulum, when released from rest.
     
  6. May 6, 2014 #5
    Right, [itex]\varphi[/itex] ranges from 0 to 2[itex]\pi[/itex]
    [itex]\vartheta[/itex] is the initial angle
     
  7. May 6, 2014 #6

    haruspex

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    No, as Doc Al posted, [itex]\alpha[/itex] is the initial angle. [itex]\vartheta[/itex] varies between [itex]-\alpha[/itex] and [itex]+\alpha[/itex].
     
  8. May 7, 2014 #7
    Maybe, i don't remember the details now
     
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