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.