For the pendulum described in \(\S 2.1\), find the equation of motion for small displacements from the position of unstable equilibrium, \(\theta=\pi\). Show that if it is released from rest at a small angle to the upward vertical, then the time taken for the angular displacement to increase by a factor of 10 will be approximately \(\sqrt{l / g} \ln 20\). Evaluate this time for a pendulum of period \(2 \mathrm{~s}\), and find the angular velocity of the pendulum when it reaches the downward vertical.

The equation of motion for a pendulum can be derived using the concept of torque about the pivot point. For small displacements, we can approximate the sine of the angle \(\theta\) to be equal to \(\theta\). Hence, considering the force acting on the pendulum, the torque about the pivot is given by: \(\tau=mgl\theta\) Here, \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(l\) is the length of the pendulum. Using Newton's second law for rotation, \(\tau=I\alpha\) where \(I=ml^2\) is the moment of inertia of the pendulum, and \(\alpha\) is the angular acceleration. So, for small displacements, \(ml^2\alpha=-mgl\theta\) Dividing the equation by \(-ml^2\) and rearranging, we get: \(\frac{d^2\theta}{dt^2}+\frac{g}{l}\theta=0\) This differential equation governs the pendulum's motion for small displacements.

To solve the differential equation \(\frac{d^2\theta}{dt^2}+\frac{g}{l}\theta=0\), we can recognize that it is a simple harmonic oscillator equation and has a solution of the form: \(\theta(t)=A\sin(\omega t+\phi)\) where \(\omega=\sqrt{\frac{g}{l}}\) is the angular frequency, \(A\) is the amplitude, and \(\phi\) is the phase angle of the oscillation. Since the pendulum is released from rest at a small angle to the upward vertical, its initial conditions are: \(\theta(0)=\theta_0\) and \(\frac{d\theta}{dt}(0)=0\) Putting these initial conditions into the general solution of the simple harmonic oscillator equation, we can determine that: \(\theta(t)=\theta_0\cos(\omega t)\) Now, we wish to find the time taken for the angular displacement to increase by a factor of 10. Therefore, we want to find the time \(t_1\) such that: \(\frac{\theta(t_1)}{\theta_0}=10\) Using the expression for \(\theta(t)\), we get: \(10\theta_0=\theta_0\cos(\omega t_1)\) This implies that \(\cos(\omega t_1)=\frac{1}{10}\). Taking the inverse cosine, we find: \(\omega t_1=\cos^{-1}\left(\frac{1}{10}\right)\) Finally, dividing by the angular frequency, we get: \(t_1=\frac{1}{\sqrt{\frac{g}{l}}}\cos^{-1}\left(\frac{1}{10}\right) \approx \sqrt{\frac{l}{g}}\ln 20\)

The angular velocity of the pendulum can be found by taking the derivative of the expression for the angular displacement with respect to time: \(\frac{d\theta}{dt}(t)=-\omega \theta_0\sin(\omega t)\) When the pendulum reaches the downward vertical, its angular displacement is given by \(\theta=\pi\), as per the problem statement. Hence, we need to find the time \(t_2\) when \(\theta(t_2)=\pi\): \(\theta_0\cos(\omega t_2)=\pi\) The time taken, \(t_2\), can be found when the pendulum reaches the downward vertical. However, it is not required to evaluate the time taken to reach this position. By directly differentiating the general equation of \(\theta(t)\) and plugging the \(\theta = \pi\) back into the expression for the angular velocity, we will directly find the angular velocity: Substituting \(\theta=\pi\) into the expression for the angular velocity, \(\omega_f=-\omega \theta_0\sin(\omega t_2)= -\omega \theta_0\sin\left(\cos^{-1}\left(\frac{-1}{\theta_0}\right)\right)\) Now, let's calculate the time for a pendulum of period \(2\mathrm{~s}\). The period is given by: \(T=\frac{2\pi}{\omega}=2\mathrm{~s}\) Using the definition of \(\omega=\sqrt{\frac{g}{l}}\), we can find the length of the pendulum: \(l=\frac{gT^2}{4\pi^2}\) Now, using the expression for \(t_1\) found in Step 2, we can evaluate the time taken for the angular displacement to increase by a factor of 10: \(t_1 \approx \sqrt{\frac{l}{g}}\ln 20\) Substituting the length of the pendulum we just found: \(t_1 \approx \sqrt{\frac{\frac{gT^2}{4\pi^2}}{g}}\ln 20 = \frac{T}{2\pi}\ln 20 \approx 0.579\mathrm{~s}\) And finally, the angular velocity of the pendulum when it reaches the downward vertical is given by: \(\omega_f = -\omega \theta_0\sin\left(\cos^{-1}\left(\frac{-1}{\theta_0}\right)\right)\)