Question

2. A simple pendulum with length $l$ satisfies the equation $\ddot{\Theta }+\frac{g}{l}sin\Theta =0$

a) If 0 is the amplitude of the oscillation, show that its period T is given by $T=\frac{4}{\omega _{0}}\int_{0}^{\pi /2}\frac{d\phi }{\sqrt{1-\alpha sin^{2}\phi }}$ where $\alpha =sin^{2}(\theta _{0}/2)$ and $\omega _{0}= \sqrt{g/l}$ .

Answer 1

Theta + w^2_0 sin theta = 0 w^2_0 = g/l or d theta/dt d^2 theta/dt^2 + d theta/dt w^2_0 sin theta = 0. d/dt [1/2 (d theta/dt)^2 - w^2_0 cos theta] = 0. Integrating on [0, t] interval. 1/2 [(d theta/dt)^2 - 1/2 ((d theta/dt)_t = 0)^2 = w^2_0 (cos theta - cos theta_0) (d theta/dt)^2 = 2 w_0 (cos theta - cos theta_0) \r\n using cos theta = 1 - 2 sin^2 theta/2 and cos theta_0 = 1 - 2 sin^2 theta_0/2 (d theta/dt)^2 = 2 w^2_0 [2 sin^2 theta_0/2 - 2 sin^2 theta/2] = 4 w^2_0 sin^2 theta/2 [1 - sin^2 theta/2/sin^2 theta_0/2] or d theta/dt = 2 w_0 sin theta_0/2 squareroot 1 - sun^2 theta/2/sin^2 theta_0/2 (1) let sin theta/2 = sin theta_0/2 sin emptyset 1/2 cos theta/2 d theta = sin (theta_0/2) cos emptyset d emptyset. d emptyset = d theta/2 w_0 sin theta_0/2 squareroot 1 - sin^2 emptyset from (1) \r\n dt = 2 sin theta_0/2 cos emptyset/cos theta/2 d emptyset/2 w_0 squareroot 1 - sin^2 emptyset middot sin theta_0/2 = d emptyset cos emptyset/w_0 cos theta/2 cos emptyset