A simple pendulum experiment is constructed from a point mass $m$ attached to a pivot by a massless rod of length $L$ in a constant gravitational field. The rod is released from an angle $\theta_0 < \frac{\pi}{2}$ at rest and the period of motion is found to be $T_0$. Ignore air resistance and friction.

What is the maximum value of the tension in the rod?

$\textbf{(A) } mg\\
\textbf{(B) }  2mg\\
\textbf{(C) } mL\theta_0/T_0^2\\
\textbf{(D) } mg \sin \theta_0\\
\textbf{(E) } mg(3 - 2 \cos \theta_0)$
1. **Identify the forces acting on the pendulum:**
   - The tension \( T \) in the rod.
   - The gravitational force \( mg \).

2. **Apply Newton's Second Law in the radial direction:**
   - The centripetal force required to keep the mass moving in a circular path is provided by the tension in the rod and the component of the gravitational force acting along the rod.
   - At any angle \(\theta\), the radial component of the gravitational force is \( mg \cos(\theta) \).
   - The centripetal acceleration is given by \( a_c = \frac{v^2}{L} \).

   Therefore, by Newton's Second Law:
   \[
   T - mg \cos(\theta) = m \frac{v^2}{L}
   \]
   Rearranging, we get:
   \[
   T = mg \cos(\theta) + m \frac{v^2}{L}
   \]

3. **Determine the velocity \( v \) at the lowest point:**
   - Use conservation of energy to find the velocity at the lowest point of the swing.
   - The potential energy at the initial height (when the pendulum is released from \(\theta_0\)) is converted into kinetic energy at the lowest point.

   The height \( h \) from the lowest point when the pendulum is at angle \(\theta_0\) is:
   \[
   h = L - L \cos(\theta_0) = L (1 - \cos(\theta_0))
   \]

   The potential energy at the initial height is:
   \[
   PE = mgh = mgL (1 - \cos(\theta_0))
   \]

   At the lowest point, all this potential energy is converted into kinetic energy:
   \[
   \frac{1}{2} mv^2 = mgL (1 - \cos(\theta_0))
   \]

   Solving for \( v^2 \):
   \[
   v^2 = 2gL (1 - \cos(\theta_0))
   \]

4. **Substitute \( v^2 \) into the tension equation:**
   \[
   T = mg \cos(\theta) + m \frac{2gL (1 - \cos(\theta_0))}{L}
   \]
   Simplifying:
   \[
   T = mg \cos(\theta) + 2mg (1 - \cos(\theta_0))
   \]

5. **Find the maximum tension:**
   - The maximum tension occurs at the lowest point of the swing, where \(\theta = 0\).
   - At \(\theta = 0\), \(\cos(0) = 1\).

   Therefore:
   \[
   T_{\text{max}} = mg \cos(0) + 2mg (1 - \cos(\theta_0))
   \]
   \[
   T_{\text{max}} = mg \cdot 1 + 2mg (1 - \cos(\theta_0))
   \]
   \[
   T_{\text{max}} = mg + 2mg - 2mg \cos(\theta_0)
   \]
   \[
   T_{\text{max}} = mg (3 - 2 \cos(\theta_0))
   \]

The final answer is \(\boxed{mg(3 - 2 \cos(\theta_0))}\).