A pendulum is made from a uniform rod of mass \(m_{1}\) and a small block of mass \(m_{2}\) attached at the lower end. (a) If the length of the pendulum is \(L\) and the oscillations are small, find the period of the oscillations in terms of \(m_{1}, m_{2}, L,\) and \(g .\) (b) Check your answer to part (a) in the two special cases \(m_{1} \gg m_{2}\) and \(m_{1}<<m_{2}\).

First, we need to compute the moment of inertia (\(I\)) for the pendulum. For a uniform rod about an axis perpendicular to it at one end, the moment of inertia is given by \(\frac{1}{3}m_1L^2\). Next, we need to consider the moment of inertia of the block of mass \(m_2\). Since the block is attached to the lower end of the rod, the moment of inertia due to the block will be \(m_2L^2\). Therefore, the total moment of inertia will be the sum of the two: \(I = \frac{1}{3}m_1L^2 + m_2L^2\).

Let's consider the pendulum at an angle \(\theta\) relative to the vertical. The restoring torque due to gravity is given by: \(\tau = -m_1g \frac{L}{2} \sin\theta - m_2g L \sin\theta\).

Since the oscillations are small, we can assume that \(\sin\theta \approx \theta\). Thus, the restoring torque is approximately \(\tau \approx - (m_1\frac{gL}{2} + m_2gL) \theta\).

According to Newton's second law in rotational form, \(\tau = I\alpha = I\frac{d^2\theta}{dt^2}\). Substitute the torque and moment of inertia expressions: \((\frac{1}{3}m_1L^2 + m_2L^2) \frac{d^2\theta}{dt^2} \approx - (m_1\frac{gL}{2} + m_2gL) \theta\).

Rewriting the equation, \(\frac{d^2\theta}{dt^2} \approx - \frac{3g(m_1 + 2m_2)}{2L(m_1 + m_2)} \theta\). The solution to this differential equation is of the form \(\theta(t) = A\cos{(\omega t + \phi)}\), where \(\omega\) is the angular frequency. Comparing the equations, we find that \(\omega^2 = \frac{3g(m_1 + 2m_2)}{2L(m_1 + m_2)}\).

The period of the oscillations, \(T\), is related to the angular frequency by \(\omega = \frac{2\pi}{T}\). Therefore, \(T = \frac{2\pi}{\omega}\). Plugging in our expression for \(\omega\), we get \(T = 2\pi\sqrt{\frac{2L(m_1+m_2)}{3g(m_1+2m_2)}}\).

(a) When \(m_1 \gg m_2\), we can simplify the expression for the period as \(T \approx 2\pi\sqrt{\frac{2L}{3g}}\), since the contribution of \(m_2\) is negligible in comparison to \(m_1\). (b) When \(m_1 \ll m_2\), the expression for the period can be simplified to \(T \approx 2\pi\sqrt{\frac{L}{g}}\), since in this case, the contribution of \(m_1\) is negligible in comparison to \(m_2\). This is the well-known formula for the period of a simple pendulum with all the mass concentrated at the bottom. In conclusion, the period of the oscillations for the given pendulum is \(T = 2\pi\sqrt{\frac{2L(m_1+m_2)}{3g(m_1+2m_2)}}\). In the special cases of \(m_1 \gg m_2\) and \(m_1 \ll m_2\), we found the approximate periods to be \(T \approx 2\pi\sqrt{\frac{2L}{3g}}\) and \(T \approx 2\pi\sqrt{\frac{L}{g}}\), respectively.