Angular SHM and Simple Pendulum: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Angular SHM and Simple Pendulum. Understanding the underlying formula is key to arriving at the correct answer below:

For the smaller pendulum, time elapsed for one oscillation is \(t = T\), so its phase is \(2\pi\). The phase of the second pendulum is \(\phi_2 = \frac{2\pi}{5T/4}T = \frac{8\pi}{5}\). The phase difference is \(2\pi - \frac{8\pi}{5} = \frac{2\pi}{5} = 72^\circ\).