Question 10.Q.2: Period of oscillation Find the period of oscillation of the ...

From the equation (10.40), it follows that the motion is restricted to those values of θ that make the right side positive, and that \dot{\theta}=0 when the right side is zero.
Hence, θ oscillates periodically in the range −π/3 < θ < π/3. Consider the first half-oscilliation. In this part of the motion, \dot{\theta}<0 and so θ satisfies the equation

\dot{\theta}^{2}=\frac{4 g}{a}\left(\frac{2 \cos \theta-1}{4-\cos ^{2} \theta}\right)            (10.40)

\dot{\theta}=-\left(\frac{4 g}{a}\right)^{1 / 2}\left(\frac{2 \cos \theta-1}{4-\cos ^{2} \theta}\right)^{1 / 2},

a first order separable ODE. On separating, we find that τ , the period of a full oscillation, is given by

\tau=\left(\frac{a}{g}\right)^{1 / 2} \int_{-\pi / 3}^{\pi / 3}\left(\frac{4-\cos ^{2} \theta}{2 \cos \theta-1}\right)^{1 / 2} \approx 6.23\left(\frac{a}{g}\right)^{1 / 2}.

Thus the determination of θ (t) has been reduced to an integration, and, with θ (t) ‘known’, the equation(10.38) can be solved to give x(t) as an integral. This confirms that the system is integrable.

4 \dot{x}+a \dot{\theta} \cos \theta=0                        (10.38)