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Homework Help: Period of pendulum motion

  1. Jan 28, 2017 #1
    1. The problem statement, all variables and given/known data

    A pendulum obeys the equation [tex] \ddot{\theta} = -\sin(\theta) [/tex] and has amplitude [tex] \theta_0 [/tex]. I have to show that the period is
    [tex] T = 4 \int_{0}^{\frac{\pi}{2}} \frac{d\phi}{\sqrt{1-\alpha \sin^2(\phi)}} [/tex] where [tex] \alpha = \sin^2(\frac{\theta_0}{2})[/tex]

    2. The attempt at a solution

    I derived an expression for time:

    [tex] \dot{\theta}\frac{d\dot{\theta}}{d\theta} = -\sin(\theta) [/tex]

    I said that the pendulum starts out at the height of its amplitude [tex] \theta = \theta_0 [/tex] where it also has zero velocity

    [tex] \int_{0}^{\dot{\theta}} \dot{\theta}d\dot{\theta} = \int_{\theta_0}^{\theta} -\sin(\theta)d\theta [/tex]

    [tex] \dot{\theta} = \frac{d\theta}{dt}= \pm \sqrt{2(\cos(\theta)-\cos(\theta_0))} [/tex]

    So for the period we can integrate from [tex] \theta_0 \text{ to } 0 [/tex], which is a quarter of the period, then multiply by 4 to get the whole period.

    [tex] T = 4 \int_{\theta_0}^{0} \frac{d\theta}{\sqrt{2(\cos(\theta)-\cos(\theta_0))}} [/tex]

    By the half-angle identity,

    [tex] T = 2 \int_{\theta_0}^{0} \frac{d\theta}{\sqrt{\sin^2(\theta_0 /2)-\sin^2(\theta /2)}} = 2 \int_{\theta_0}^{0} \frac{d\theta}{\sqrt{\alpha-\sin^2(\theta /2)}} [/tex]

    And this is where I'm stuck. It looks similar to the desired answer, but I can't think of any identity or substitution that would give the right integrand and limits.
    Last edited: Jan 28, 2017
  2. jcsd
  3. Jan 29, 2017 #2


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