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.

The experiment is repeated with a new pendulum with a rod of length $4L$, using the same angle $\theta_0$, and the period of motion is found to be $T$. Which of the following statements is correct?

$\textbf{(A) } T = 2T_0 \text{ regardless of the value of } \theta_0\\
\textbf{(B) }  T > 2T_0 \text{ with } T \approx 2T_0 \text{ if } \theta_0 \ll 1\\
\textbf{(C) }  T < 2T_0 \text{ with } T \approx 2T_0 \text{ if } \theta_0 \ll 1\\
\textbf{(D) } T < 2T_0 \text{ with some values of } \theta_0 \text{ and } T > 2T_0 \text{ for other values of } \theta_0\\
\textbf{(E) } T \text{ and } T_0 \text{ are not defined because the motion is not periodic unless } \theta_0 \ll 1$
1. The period of a simple pendulum for small angles $\theta_0$ is given by the formula:
   \[
   T = 2\pi \sqrt{\frac{L}{g}}
   \]
   where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.

2. For the initial pendulum with length $L$, the period is:
   \[
   T_0 = 2\pi \sqrt{\frac{L}{g}}
   \]

3. When the length of the pendulum is increased to $4L$, the new period $T$ is:
   \[
   T = 2\pi \sqrt{\frac{4L}{g}} = 2\pi \sqrt{4} \sqrt{\frac{L}{g}} = 2 \cdot 2\pi \sqrt{\frac{L}{g}} = 2T_0
   \]

4. This result holds true for small angles $\theta_0 \ll 1$, where the approximation $\sin(\theta) \approx \theta$ is valid. For larger angles, the period depends on the integral form which involves the elliptic integral, making the exact function $f(\theta_0)$ more complex.

5. However, for small angles, the period is approximately:
   \[
   T \approx 2T_0
   \]

6. Therefore, the correct statement is:
   \[
   \boxed{\textbf{(A) } T = 2T_0 \text{ regardless of the value of } \theta_0}
   \]