91Ó°ÊÓ

Let's consider \(\theta\) as the angular displacement of the rod from the equilibrium position. The torque about the pivot due to gravity is \(m \cdot g \cdot L/2 \cdot sin(\theta)\), where \(L\) is the length of the rod, \(m\) is its mass, and \(g\) is the acceleration due to gravity. For small oscillations, \(sin(\theta)\) can be approximated as \(\theta\). Therefore, the torque is \(m \cdot g \cdot L/2 \cdot \theta\). According to Newton's second law for rotation, the torque is also equal to \(I \cdot d²\theta/dt²\), where \(I\) is the moment of inertia of the rod about the pivot point, equal to \(m \cdot L²/3\). Equating these two expressions and rearranging, we get \(d²\theta/dt² + 3g/(2L) \cdot \theta = 0\), which is the differential equation for small oscillations.

A system is underdamped if the damping coefficient \(c\) is less than the quotient of the square root of the mass \(m\) and the spring constant \(k\) divided by 2. For an oscillating rod, the effective mass is \(m=L/3\), and the effective spring constant is \(k=m \cdot g/L\). Substituting these values, we get \(c=\sqrt(g \cdot L/3)/2\). Therefore, if \(c<\sqrt{m k} / 2\), then the system remains underdamped.