Imagine you are an astronaut who has landed on another planet and wants to determine the free-fall acceleration on that planet. In one of the experiments you decide to conduct, you use a pendulum 0.47 m long and find that the period of oscillation for this pendulum is 1.43 s. What is the acceleration due to gravity on that planet? (Enter the magnitude only.)

I know the formula to use is
T=2pi(L/g)^0.5(g) but I keep getting a wrong answer

t^2=2x3.14x0.47/a

Or, a=6.28x0.47/1.43^2=0.43 m/sec^2

To determine the acceleration due to gravity on the planet using a pendulum, you need to rearrange the formula for the period of oscillation. The correct formula is:

T = 2π * √(L/g)

Where:
T is the period of oscillation (given as 1.43 s).
π is a mathematical constant (approximately 3.14159).
L is the length of the pendulum (given as 0.47 m).
g is the acceleration due to gravity (what we're trying to find).

Now, let's solve the equation for g:

T = 2π * √(L/g)

Rearrange the equation to isolate g:

T/2π = √(L/g)

Square both sides of the equation to remove the square root:

(T/2π)^2 = (L/g)

Simplify the right side of the equation:

g = L / (T/2π)^2

Substitute the given values:

g = 0.47 / (1.43/2π)^2

Now, calculate the value of g:

g = 0.47 / (1.43/(2*3.14159))^2

g ≈ 3.44 m/s²

Therefore, the acceleration due to gravity on the planet is approximately 3.44 m/s².