Physics
posted by Ray on .
Imagine you are an astronaut who has landed on another planet and wants to determine the freefall 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 pi sqrt(L/g)
1.4 = 6.28 sqrt (.47/g)
.47/g = .0497
g = 9.46 m/s^2 
It still says its the wrong answer. Let me calculate it again

That assumes that you know about simple pendula
small angle A at top, g and L and m
KE at bottom = (1/2) m V^2
V is max speed at bottom = L d angle/dt
Pe at top = m g h = m g L(1 cos A)
for small A, cos A = 1  A^2/2 + ....
m g L (A^2/2) = (1/2) m V^2
g L A^2 = V^2
now assume sinusoidal motion
angle = A sin (2 pi/T) t
d angle/dt = (2 pi/T) A cos (2 pi/T)t
x max = A
v max = L d angle/dt = (2 pi L/T)A
then
g L A^2 = (2 pi L/T)^2 A^2
g/L = (2 pi/T)^2
T^2 = (2 pi)^2 L/g
T = 2 pi sqrt (L/g) 
I used .14 not .143

1.43 = 6.28 sqrt(.47/g)
try 9.06 m/s^2 
For some reason its still saying its the wrong answer

Your saying g=9.06

yes

Yes, it finally worked thank you so much.