Solution:

Given the SHM equation:

\[
2 \frac{d^2x}{dt^2} + 32x = 0
\]

We can rewrite it as:

\[
\frac{d^2x}{dt^2} + 16x = 0
\]

This equation is of the form:

\[
\frac{d^2x}{dt^2} + \omega^2 x = 0
\]

where \( \omega^2 = 16 \).

Solution:

1. Angular Frequency \( \omega \):
\[
\omega = \sqrt{16} = 4 \, \text{rad/s}
\]

2. Time Period \( T \):
\[
T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{s}
\]

 Answer:
The time period of the oscillation is \( \frac{\pi}{2} \) seconds.