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In the pendulum formula [tex]T=2 \pi \sqrt{\frac{L}{g}}[/tex], we use [tex]g=9.8 \, \text{m/s}^2[/tex] for the acceleration due to gravity on Earth. But what about on the Moon? If an astronaut on the surface of the Moon swings a 1-meter long pendulum, and it has a period of 4.9 seconds, what is the acceleration due to gravity, [tex]g[/tex], on the Moon?

A. [tex]0.7 \, \text{m/s}^2[/tex]
B. [tex]2.8 \, \text{m/s}^2[/tex]
C. [tex]1.2 \, \text{m/s}^2[/tex]
D. [tex]1.6 \, \text{m/s}^2[/tex]



Answer :

GinnyAnswer
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]

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