To find the acceleration due to gravity, \( g \), on the Moon, we'll use the pendulum formula:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{g}} \][/tex]
Given:
- The period \( T = 4.9 \) seconds
- The length \( L = 1.0 \) meter
We need to solve for \( g \). Rearranging the formula to isolate \( g \) gives:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{g}} \][/tex]
Square both sides to get rid of the square root:
[tex]\[ T^2 = (2 \pi)^2 \frac{L}{g} \][/tex]
[tex]\[ T^2 = 4 \pi^2 \frac{L}{g} \][/tex]
Now solve for \( g \):
[tex]\[ g = \frac{4 \pi^2 L}{T^2} \][/tex]
Substitute the known values into the equation:
[tex]\[ g = \frac{4 \pi^2 \times 1.0 \, \text{m}}{(4.9 \, \text{s})^2} \][/tex]
Upon calculating this, we find that:
[tex]\[ g \approx 1.644 \, \text{m/s}^2 \][/tex]
Therefore, the acceleration due to gravity on the Moon is approximately:
[tex]\[ g \approx 1.6 \, \text{m/s}^2 \][/tex]
Hence, the correct answer is:
D. [tex]\( 1.6 \, m/s^2 \)[/tex]
