The frequency \( f \) of a simple pendulum is:

\(f = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \quad \text{where } L = 1.2 \, \text{m}, \, g = 10 \, \text{m/s}^2, \, \pi \approx \frac{22}{7}\)

Calculate:

\(\frac{g}{L} = \frac{10}{1.2} = \frac{25}{3}, \quad \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}}\)

\(f = \frac{1}{2 \times \frac{22}{7}} \times \frac{5}{\sqrt{3}} = \frac{35}{44\sqrt{3}}\)

Approximating \( \sqrt{3} \approx 1.732 \):

\(44 \times 1.732 \approx 76.208 \quad \Rightarrow \quad \frac{35}{76.208} \approx 0.459 \, \text{Hz}\)

Thus, the answer is \( D. \, 0.5 \) Hz.